鏍?wèi)鐘舵暟缁勫Q?榪樹(shù)笉浼?xì)鐢ㄥQ屽嚑涔庣収鎶勫埆浜虹殑
#include<iostream>
#include<stdio.h>
using namespace std;
const int MAX_X=32001,MAX_N=15000;
int c[MAX_X+1]={0},sum[MAX_N+1]={0};
int n;
int lowbit(int x){ return x&(-x); }
void inc(int x)
{
while(x<=MAX_X)
{
c[x]++;
x+=lowbit(x);
}
}
int Sum(int x)
{
int ans=0;
while(x>=1)
{
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
int x,y,i;
cin>>n;
for(i=1; i<=n; i++)
{
cin>>x>>y;
x++; y++;
sum[Sum(x)]++;
inc(x);
}
for(i=0; i<n; i++)
printf("%d\n",sum[i]);//cout<<sum[i]<<endl;
system("pause");
return 0;
}
]]>
1060. Flip Game
Time Limit: 2.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
Flip game is played on a rectangular 4×4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:
- Choose any one of the 16 pieces.
- Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).
Consider the following position as an example:
bwbw
wwww
bbwb
bwwb
|
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:
bwbw
bwww
wwwb
wwwb
|
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.
Input
The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.
Output
Write to the output a single integer number 鈥?the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).
Sample
| input | output |
|---|---|
bwbw
wwww
bbwb
bwwb
|
Impossible
|
Problem Source: 2000-2001 ACM Northeastern European Regional Programming Contest
wa浜?jiǎn)鍑爧?br>鍘熷洜錛?鏄痗nt鍙娌″噺
2鏄痵earch鐨勬帹鍑烘潯浠跺啓鎴恑f(m==16){ ……}錛岃繖縐嶆儏鍐典笅瀵逛簬鍙敼鍙樻渶鍚庝竴涓嵆鍙緱鎯呭喌娌℃湁鍒ゅ畾銆傛敼鎴恗>16
#include<iostream>
using namespace std;
bool status[6][6]={0};
const int MAX=0x7fffffff;
int cnt=0,cntMin=MAX;
bool check()
{
bool f=status[1][1];
int i,j;
for(i=1; i<=4; i++)
for(j=1; j<=4; j++)
if(f!=status[i][j])return false;
return true;
}
void turn(int i, int j)
{
status[i][j]=!status[i][j];
status[i-1][j]=!status[i-1][j];
status[i+1][j]=!status[i+1][j];
status[i][j-1]=!status[i][j-1];
status[i][j+1]=!status[i][j+1];
}
void input()
{
char ch;
for(int i=1; i<=4; i++)
for(int j=1; j<=4; j++)
{
cin>>ch;
status[i][j]=(ch=='b' ? 0 : 1 );
}
}
void search(int m)
{
if(m>16){
if( check() && (cnt<cntMin) )cntMin=cnt;
return ;
}
int i=(m+3)/4;
int j=m-4*(i-1);
turn(i,j); cnt++;
search(m+1);
turn(i,j);cnt--;
search(m+1);
}
int main()
{
input();
search(1);
if(cntMin==MAX)cout<<"Impossible"<<endl;
else cout<<cntMin<<endl;
system("pause");
return 0;
}
using namespace std;
bool status[6][6]={0};
const int MAX=0x7fffffff;
int cnt=0,cntMin=MAX;
bool check()
{
bool f=status[1][1];
int i,j;
for(i=1; i<=4; i++)
for(j=1; j<=4; j++)
if(f!=status[i][j])return false;
return true;
}
void turn(int i, int j)
{
status[i][j]=!status[i][j];
status[i-1][j]=!status[i-1][j];
status[i+1][j]=!status[i+1][j];
status[i][j-1]=!status[i][j-1];
status[i][j+1]=!status[i][j+1];
}
void input()
{
char ch;
for(int i=1; i<=4; i++)
for(int j=1; j<=4; j++)
{
cin>>ch;
status[i][j]=(ch=='b' ? 0 : 1 );
}
}
void search(int m)
{
if(m>16){
if( check() && (cnt<cntMin) )cntMin=cnt;
return ;
}
int i=(m+3)/4;
int j=m-4*(i-1);
turn(i,j); cnt++;
search(m+1);
turn(i,j);cnt--;
search(m+1);
}
int main()
{
input();
search(1);
if(cntMin==MAX)cout<<"Impossible"<<endl;
else cout<<cntMin<<endl;
system("pause");
return 0;
}
]]>
1080. Map Colouring
Time Limit: 1.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
We consider a geographical map with N countries numbered from 1 to N (0 < N < 99). For every country we know the numbers of other countries which are connected with its border. From every country we can reach to any other one, eventually crossing some borders. Write a program which determines whether it is possible to colour the map only in two colours 鈥?red and blue in such a way that if two countries are connected their colours are different. The colour of the first country is red. Your program must output one possible colouring for the other countries, or show, that such colouring is impossible.
Input
On the first line is written the number N. On the following N lines, the i-th line contains the countries to which the i-th country is connected. Every integer on this line is bigger than i, except the last one which is 0 and marks that no more countries are listed for country i. If a line contains 0, that means that the i-th country is not connected to any other country, which number is larger than i.
Output
The output contains exactly one line. If the colouring is possible, this line must contain a list of zeros and ones, without any separators between them. The i-th digit in this sequence is the colour of the i-th country. 0 corresponds to red colour, and one 鈥?to blue colour. If a colouring is not possible, output the integer −1.
Sample
| input | output |
|---|---|
3
2 0
3 0
0
|
010
|
DFS:鎴朆FS錛屾垨鑰呭茍鏌ラ泦錛堜笉浼?xì)鐢ㄥQ?/pre>榪欓噷鐢ㄧ殑DFS錛岀敤涓涓爣璁版暟緇勶紝娌¤繘鍏ヤ竴涓仈閫氬垎鍥懼悗錛屾爣璁頒負(fù)0錛堣〃紺轟竴縐嶉鑹詫級(jí)涓庡畠鐩歌繛鐨勬爣璁頒負(fù)1錛堝彟涓縐嶉鑹詫級(jí)錛?/pre>鐒跺悗涓?鐩歌繛鐨勫湪鏍囪涓?銆?榪欓噷鐨勬爣璁伴兘鏄娌℃湁鏍囪榪囩殑榪涜鐨勶紝濡傛灉鏄爣璁拌繃鐨勶紝灝辮媯(gè)鏌ヤ粬浠殑鏍囪鏄惁鐩稿悓濡傛灉鐩稿悓鍒欒鏄庯紝浠栦滑鍚岃壊銆?br>//ural 1080
#include<iostream>
using namespace std;
const int MAX=100;
bool adj[MAX][MAX];
int flg[MAX];
int n;
bool isPossible=true;
void input()
{
cin>>n;
int temp;
for(int i=1; i<=n; i++)
{
while(cin>>temp,temp!=0)
{
adj[i][temp]=adj[temp][i]=true;
}
}
}
void dfs(int i)
{
if(isPossible==false)return ;
if(flg[i]==-1)flg[i]=0;
for(int j=1; j<=n; j++)
{
if(adj[i][j])
{
if(flg[j]==-1){ flg[j]=flg[i]==0? 1:0; dfs(j); }
else if(flg[j]==flg[i])isPossible=false;
}
}
}
int main()
{
memset(adj,0,sizeof adj);
input();
for(int i=1; i<=n; i++)
flg[i]=-1;
flg[1]=0;
for(int i=1; i<=n; i++)
dfs(i);
if(isPossible==false)cout<<-1<<endl;
else {
for(int k=1; k<=n; k++)
cout<<flg[k];
cout<<endl;
}
system("pause");
return 0;
}
]]>
1008. Image encoding
Time Limit: 2.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
There are several ways to encode an image. In this problem we will consider two representations of an image. We assume that the image consists of black and white pixels. There is at least one black pixel and all black pixels are connected with their sides. Coordinates of black pixels are not less than 1 and not greater than 10. An example of such an image is at the figure.
Both representations describe arrangement of black pixels only.
At the first representation we specify in the first line number of black pixels and coordinates of each black pixel in the following lines. Pixels are listed in order of increasing X. In case of equality of X they are listed in order of increasing Y. Image at the figure is encoded as follows:
6
2 3
2 4
3 3
3 4
4 2
4 3
2 3
2 4
3 3
3 4
4 2
4 3
At the second representation we specify in the first line coordinates of the lowest left black pixel. Each of the following lines contains a description of neighbors for one of the pixels. At first, neighbors of the lowest left pixel are specified, then neighbors of its first neighbor (if it exists) are specified, then neighbors of its second neighbor (if it also exists) follow. When all its neighbors are described the description of the neighbors of its first neighbor follows. The description of the neighbors of its second neighbor follows then and so on.
Each descriptive line contains at most one letter for each neighbor: R for the right, T for the top, L for the left, B for the bottom. If the neighbor was already specified it is not included into the descriptive line and vice-versa. Also there is only one descriptive line for each pixel. Neighbors are listed counter-clockwise starting with the right. Each descriptive line except the last ends with a comma. The last line ends with a full stop. Image at the figure is encoded as follows:
2 3
RT,
RT,
,
B,
,
.
RT,
RT,
,
B,
,
.
There are no leading or tailing spaces in any representation. There is exactly one space between X and Y coordinates.
Input
One representation of the image will be given to your program in the input.
Output
Your program has to write other representation of the image to the output.
Sample
| input | output |
|---|---|
6
2 3
2 4
3 3
3 4
4 2
4 3
|
2 3
RT,
RT,
,
B,
,
.
|
Problem Source: Third Open USTU Collegiate Programming Contest (PhysTech Cup), March 18, 2000
榪欓鑺變簡(jiǎn)涓嶅皯鏃墮棿錛岄鐩病鐪嬫竻錛屽彧鐪嬬ず渚嬩互涓哄彧瑕佷粠絎竴縐嶆柟妗堣漿鎹㈡垚絎簩縐嶏紝娌℃兂鍒拌繕鏈変粠絎簩縐嶈漿鎹㈡垚絎竴縐?br>BFS錛?
| Accepted | |
0.015 | 217 KB |
#include<iostream>
#include<queue>
#include<string.h>
#include<string>
using namespace std;
typedef struct point
{int x,y; } point;
bool graph[12][12]={0};
bool f[12][12]={0};
bool print[12][12]={0};
int n;
int lbx=10,lby=10;
void input()
{
int x,y;
for(int i=1; i<=n; i++)
{
cin>>x>>y; graph[x][y]=1;
if(x<lbx){ lbx=x; lby=y; }
else if(x==lbx&&y<lby)
lby=y;
}
}
void BFS1() //鏁板瓧鎯呭喌
{
queue<point> q;
point temp; temp.x=lbx; temp.y=lby;
cout<<lbx<<' '<<lby<<endl;
q.push(temp);
int x,y; int cnt=0;
print[lbx][lby]=1;
while(q.size())
{
temp=q.front(); q.pop();
x=temp.x; y=temp.y;
if(f[x][y]==1)continue;
f[x][y]=1;
if(graph[x+1][y]==1&&!f[x+1][y])
{if(!print[x+1][y]) cout<<'R'; print[x+1][y]=1; temp.x=x+1;temp.y=y; q.push(temp);}
if(graph[x][y+1]==1&&!f[x][y+1])
{if(!print[x][y+1]) cout<<'T'; print[x][y+1]=1; temp.x=x;temp.y=y+1; q.push(temp);}
if(graph[x-1][y]==1&&!f[x-1][y])
{if(!print[x-1][y]) cout<<'L'; print[x-1][y]=1; temp.x=x-1;temp.y=y; q.push(temp);}
if(graph[x][y-1]==1&&!f[x][y-1])
{if(!print[x][y-1]) cout<<'B'; print[x][y-1]=1; temp.x=x;temp.y=y-1; q.push(temp);}
++cnt;
if(cnt==n) cout<<'.'<<endl;
else cout <<','<<endl;
}
}
void BFS2() // 浠嶳 T L B鎻忚堪杞崲鎴愬潗鏍?/span>
{
lbx=n; cin>>lby;
queue<point>q;
point temp; temp.x=lbx; temp.y=lby;
q.push(temp);
string str; int x,y;
graph[lbx][lby]=1;
while(q.size())
{
temp=q.front(); q.pop();
x=temp.x; y=temp.y;
if(f[x][y])continue;
f[x][y]=1;
cin>>str;
for(int i=0; i<str.size()-1; i++)
{
if(str[i]=='R') {graph[x+1][y]=1; if(!f[x+1][y]){ temp.x=x+1; temp.y=y; q.push(temp);} }
else if(str[i]=='T'){graph[x][y+1]=1; if(!f[x][y+1]){ temp.x=x; temp.y=y+1; q.push(temp);} }
else if(str[i]=='L'){graph[x-1][y]=1; if(!f[x-1][y]){ temp.x=x-1; temp.y=y; q.push(temp);} }
else if(str[i]=='B'){graph[x][y-1]=1;if(!f[x][y-1]){ temp.x=x; temp.y=y-1; q.push(temp);} }
}
if(str[str.size()-1]=='.')break;
}
int cnt=0;
for(int i=1; i<=10; i++)
for(int j=1; j<=10; j++)
if(graph[i][j])cnt++;
cout<<cnt<<endl;
for(int i=1; i<=10; i++)
for(int j=1; j<=10; j++)
if(graph[i][j])cout<<i<<' '<<j<<endl;
}
int main()
{
cin>>n;
if(getchar()=='\n')
{
input();
BFS1();
}
else BFS2();
system("pause");
return 0;
}
#include<queue>
#include<string.h>
#include<string>
using namespace std;
typedef struct point
{int x,y; } point;
bool graph[12][12]={0};
bool f[12][12]={0};
bool print[12][12]={0};
int n;
int lbx=10,lby=10;
void input()
{
int x,y;
for(int i=1; i<=n; i++)
{
cin>>x>>y; graph[x][y]=1;
if(x<lbx){ lbx=x; lby=y; }
else if(x==lbx&&y<lby)
lby=y;
}
}
void BFS1() //鏁板瓧鎯呭喌
{
queue<point> q;
point temp; temp.x=lbx; temp.y=lby;
cout<<lbx<<' '<<lby<<endl;
q.push(temp);
int x,y; int cnt=0;
print[lbx][lby]=1;
while(q.size())
{
temp=q.front(); q.pop();
x=temp.x; y=temp.y;
if(f[x][y]==1)continue;
f[x][y]=1;
if(graph[x+1][y]==1&&!f[x+1][y])
{if(!print[x+1][y]) cout<<'R'; print[x+1][y]=1; temp.x=x+1;temp.y=y; q.push(temp);}
if(graph[x][y+1]==1&&!f[x][y+1])
{if(!print[x][y+1]) cout<<'T'; print[x][y+1]=1; temp.x=x;temp.y=y+1; q.push(temp);}
if(graph[x-1][y]==1&&!f[x-1][y])
{if(!print[x-1][y]) cout<<'L'; print[x-1][y]=1; temp.x=x-1;temp.y=y; q.push(temp);}
if(graph[x][y-1]==1&&!f[x][y-1])
{if(!print[x][y-1]) cout<<'B'; print[x][y-1]=1; temp.x=x;temp.y=y-1; q.push(temp);}
++cnt;
if(cnt==n) cout<<'.'<<endl;
else cout <<','<<endl;
}
}
void BFS2() // 浠嶳 T L B鎻忚堪杞崲鎴愬潗鏍?/span>
{
lbx=n; cin>>lby;
queue<point>q;
point temp; temp.x=lbx; temp.y=lby;
q.push(temp);
string str; int x,y;
graph[lbx][lby]=1;
while(q.size())
{
temp=q.front(); q.pop();
x=temp.x; y=temp.y;
if(f[x][y])continue;
f[x][y]=1;
cin>>str;
for(int i=0; i<str.size()-1; i++)
{
if(str[i]=='R') {graph[x+1][y]=1; if(!f[x+1][y]){ temp.x=x+1; temp.y=y; q.push(temp);} }
else if(str[i]=='T'){graph[x][y+1]=1; if(!f[x][y+1]){ temp.x=x; temp.y=y+1; q.push(temp);} }
else if(str[i]=='L'){graph[x-1][y]=1; if(!f[x-1][y]){ temp.x=x-1; temp.y=y; q.push(temp);} }
else if(str[i]=='B'){graph[x][y-1]=1;if(!f[x][y-1]){ temp.x=x; temp.y=y-1; q.push(temp);} }
}
if(str[str.size()-1]=='.')break;
}
int cnt=0;
for(int i=1; i<=10; i++)
for(int j=1; j<=10; j++)
if(graph[i][j])cnt++;
cout<<cnt<<endl;
for(int i=1; i<=10; i++)
for(int j=1; j<=10; j++)
if(graph[i][j])cout<<i<<' '<<j<<endl;
}
int main()
{
cin>>n;
if(getchar()=='\n')
{
input();
BFS1();
}
else BFS2();
system("pause");
return 0;
}
]]>
1048. Superlong sums
Time Limit: 2.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
The creators of a new programming language D++ have found out that whatever limit for SuperLongInt type they make, sometimes programmers need to operate even larger numbers. A limit of 1000 digits is so small… You have to find the sum of two numbers with maximal size of 1 000 000 digits.
Input
The first line contains a single number N (1 ≤ N ≤ 1 000 000) 鈥?the length of the integers (in order to make their lengths equal, some leading zeroes can be added). It is followed by these integers written in columns. That is, the next N lines contain two digits each, divided by a space. Each of the two given integers is not less than 1, and the length of their sum does not exceed N.
Output
Output exactly N digits in a single line representing the sum of these two integers.
Sample
| input | output |
|---|---|
4
0 4
4 2
6 8
3 7
|
4750 |
Problem Author: Stanislav Vasilyev and Alexander Klepinin
Problem Source: Ural State University collegiate programming contest (25.03.2000)
Problem Source: Ural State University collegiate programming contest (25.03.2000)
綆鍗曞ぇ鏁板姞娉?br>
#include<iostream>
using namespace std;
int a[1000005]={0};
int b[1000005]={0};
int ans[1000005]={0};
int main()
{
int N,i,c=0;
cin>>N;
for(i=1; i<=N; i++)
cin>>a[i]>>b[i];
for(i=N; i>=1; i--)
{
ans[i]=(a[i]+b[i]+c)%10;
c=(a[i]+b[i]+c)/10;
}
for(i=1 ; i<=N; i++)
cout<<ans[i];
cout<<endl;
system("pause");
return 0;
}
using namespace std;
int a[1000005]={0};
int b[1000005]={0};
int ans[1000005]={0};
int main()
{
int N,i,c=0;
cin>>N;
for(i=1; i<=N; i++)
cin>>a[i]>>b[i];
for(i=N; i>=1; i--)
{
ans[i]=(a[i]+b[i]+c)%10;
c=(a[i]+b[i]+c)/10;
}
for(i=1 ; i<=N; i++)
cout<<ans[i];
cout<<endl;
system("pause");
return 0;
}
]]>
1047. Simple calculations
Time Limit: 1.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
There is a sequence of N + 2 elements a0, a1, …, aN+1 (1 ≤ N ≤ 3000, −2000 ≤ ai ≤ 2000). It is known that
ai = (ai−1 + ai+1)/2 − ci
for each i = 1, 2, …, N.
You are given a0, aN+1, c1, …, cN. Write a program which calculates a1.
Input
The first line contains an integer N. The next two lines consist of numbers a0 and aN+1 each having two digits after decimal point, and the next N lines contain numbers ci (also with two digits after decimal point), one number per line.
Output
Output a1 in the same format as a0 and aN+1.
Sample
| input | output |
|---|---|
1
50.50
25.50
10.15
|
27.85
|
Problem Author: Dmitry Filimonenkov
Problem Source: Ural State University collegiate programming contest (25.03.2000)
Problem Source: Ural State University collegiate programming contest (25.03.2000)
鐢ㄦ暟瀛︽柟娉曟帹鍑哄叕寮忓嵆鍙細(xì)
//ural1047
//a1=( aN_1 +N*a0 -2*( (c1+……cN)+ (c1+……cN-1 ) +……(c1+c2)+(c1) ))/(N+1)
#include<iostream>
using namespace std;
double ci[3010]={0}; //ci[i]=c1+c2+……ci
int main()
{
int N,i;
double a0,aN_1,temp,sum=0;
cin>>N>>a0>>aN_1;
for(i=1; i<=N; i++)
{
cin>>temp;
ci[i]=ci[i-1]+temp;
sum+=ci[i];
}
sum*=2;
cout.precision(2);
cout<<fixed<<( aN_1+ N*a0 -sum)/(N+1)<<endl;
system("pause");
return 0;
}
//a1=( aN_1 +N*a0 -2*( (c1+……cN)+ (c1+……cN-1 ) +……(c1+c2)+(c1) ))/(N+1)
#include<iostream>
using namespace std;
double ci[3010]={0}; //ci[i]=c1+c2+……ci
int main()
{
int N,i;
double a0,aN_1,temp,sum=0;
cin>>N>>a0>>aN_1;
for(i=1; i<=N; i++)
{
cin>>temp;
ci[i]=ci[i-1]+temp;
sum+=ci[i];
}
sum*=2;
cout.precision(2);
cout<<fixed<<( aN_1+ N*a0 -sum)/(N+1)<<endl;
system("pause");
return 0;
}
]]>
1033. Labyrinth
Time Limit: 1.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
Administration of the labyrinth has decided to start a new season with new wallpapers. For this purpose they need a program to calculate the square of the walls inside the labyrinth. This job is just for you!
The labyrinth is represented by a matrix N×N (3 ≤ N ≤ 33, you see, ‘3’ is a magic digit!). Some matrix cells contain a dot character (‘.’) that denotes an empty square. Other cells contain a diesis character (‘#’) that denotes a square filled by monolith block of stone wall. All squares are of the same size 3×3 meters.
The walls are constructed around the labyrinth (except for the upper left and lower right corners, which are used as entrances) and on the cells with a diesis character. No other walls are constructed. There always will be a dot character at the upper left and lower right corner cells of the input matrix.
Your task is to calculate the square of visible part of the walls inside the labyrinth. In other words, the square of the walls' surface visible to a visitor of the labyrinth. Note that there's no holes to look or to move through between any two adjacent blocks of the wall. The blocks are considered to be adjacent if they touch each other in any corner. See picture for an example: visible walls inside the labyrinth are drawn with bold lines. The height of all the walls is 3 meters.
Input
The first line of the input contains the single number N. The next N lines contain N characters each. Each line describes one row of the labyrinth matrix. In each line only dot and diesis characters will be used and each line will be finished with a new line character. There will be no spaces in the input.
Output
Your program should print to the output a single integer 鈥?the exact value of the square of the wallpaper needed.
Sample
| input | output |
|---|---|
5
.....
...##
..#..
..###
.....
|
198
|
鎼滅儲(chǔ)棰橈細(xì)娉ㄦ剰鐨勮浠庝袱涓叆鍙ｅ鎼滅儲(chǔ)錛岄槻姝腑闂存柇寮浜?br>wa浜?jiǎn)涓?鎵句笉鍑洪敊璇紝search(i+1錛宩)鍐欐垚search(i+1,j+1)錛?br>
#include<iostream>
#include<cstring>
using namespace std;
int const maxSize=35;
class ural1033
{
public:
ural1033(){ size=0; memset(f,0,sizeof f); }
void input();
void print();
void search(int i, int j);
int size;
int getn(){return N;}
private:
char a[maxSize][maxSize];
bool f[maxSize][maxSize];
int N;
};
void ural1033::input()
{
cin>>N;
int i,j;
for(i=1; i<=N; i++)
for(j=1; j<=N; j++)
cin>>a[i][j];
for(i=2; i<=N+1; i++)a[0][i]='#';
for(i=2; i<=N+1; i++)a[i][0]='#';
for(i=1; i<=N-1; i++)a[N+1][i]='#';
for(i=1; i<=N-1; i++)a[i][N+1]='#';
}
void ural1033:: search(int i, int j)
{
if(i<1||i>N||j<1||j>N||a[i][j]=='#'||f[i][j]==1)return ;
f[i][j]=1;
if(a[i-1][j]=='#')size++;
else search(i-1,j);
if(a[i][j-1]=='#')size++;
else search(i,j-1);
if(a[i][j+1]=='#')size++;
else search(i,j+1);
if(a[i+1][j]=='#')size++;
else search(i+1,j);
}
void ural1033::print()
{
for(int i=0; i<=N+1; i++,cout<<endl)
for(int j=0; j<=N+1; j++)
cout<<a[i][j]<<' ';
cout<<endl<<endl;
for(int i=0; i<=N+1; i++,cout<<endl)
for(int j=0; j<=N+1; j++)
cout<<f[i][j]<<' ';
}
int main()
{
ural1033 ural;
ural.input();
ural.search(1,1);
ural.search(ural.getn(),ural.getn());
cout<<ural.size*9<<endl;
system("pause");
return 0;
}
#include<cstring>
using namespace std;
int const maxSize=35;
class ural1033
{
public:
ural1033(){ size=0; memset(f,0,sizeof f); }
void input();
void print();
void search(int i, int j);
int size;
int getn(){return N;}
private:
char a[maxSize][maxSize];
bool f[maxSize][maxSize];
int N;
};
void ural1033::input()
{
cin>>N;
int i,j;
for(i=1; i<=N; i++)
for(j=1; j<=N; j++)
cin>>a[i][j];
for(i=2; i<=N+1; i++)a[0][i]='#';
for(i=2; i<=N+1; i++)a[i][0]='#';
for(i=1; i<=N-1; i++)a[N+1][i]='#';
for(i=1; i<=N-1; i++)a[i][N+1]='#';
}
void ural1033:: search(int i, int j)
{
if(i<1||i>N||j<1||j>N||a[i][j]=='#'||f[i][j]==1)return ;
f[i][j]=1;
if(a[i-1][j]=='#')size++;
else search(i-1,j);
if(a[i][j-1]=='#')size++;
else search(i,j-1);
if(a[i][j+1]=='#')size++;
else search(i,j+1);
if(a[i+1][j]=='#')size++;
else search(i+1,j);
}
void ural1033::print()
{
for(int i=0; i<=N+1; i++,cout<<endl)
for(int j=0; j<=N+1; j++)
cout<<a[i][j]<<' ';
cout<<endl<<endl;
for(int i=0; i<=N+1; i++,cout<<endl)
for(int j=0; j<=N+1; j++)
cout<<f[i][j]<<' ';
}
int main()
{
ural1033 ural;
ural.input();
ural.search(1,1);
ural.search(ural.getn(),ural.getn());
cout<<ural.size*9<<endl;
system("pause");
return 0;
}
]]>
#include<iostream>
using namespace std;
void f2()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
if(i1==i2)count++;
cout<<count<<endl;
}
void f4()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
if(i1+i2==i3+i4)count++;
cout<<count<<endl;
}
void f6()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
for(int i5=0; i5<=9; i5++)
for(int i6=0; i6<=9; i6++)
if(i1+i2+i3==i5+i4+i6)count++;
cout<<count<<endl;
}
void f8()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
for(int i5=0; i5<=9; i5++)
for(int i6=0; i6<=9; i6++)
for(int i7=0; i7<=9; i7++)
for(int i8=0; i8<=9; i8++)
if(i1+i2+i3+i4==i5+i7+i6+i8)count++;
cout<<count<<endl;
}
int main()
{
int n;
cin>>n;
if(n==2)f2();
else if(n==4)f4();
else if(n==6)f6();
else if(n==8)f8();
system("pause");
return 0;
}
using namespace std;
void f2()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
if(i1==i2)count++;
cout<<count<<endl;
}
void f4()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
if(i1+i2==i3+i4)count++;
cout<<count<<endl;
}
void f6()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
for(int i5=0; i5<=9; i5++)
for(int i6=0; i6<=9; i6++)
if(i1+i2+i3==i5+i4+i6)count++;
cout<<count<<endl;
}
void f8()
{
int count=0;
for(int i1=0; i1<=9; i1++)
for(int i2=0; i2<=9; i2++)
for(int i3=0; i3<=9; i3++)
for(int i4=0; i4<=9; i4++)
for(int i5=0; i5<=9; i5++)
for(int i6=0; i6<=9; i6++)
for(int i7=0; i7<=9; i7++)
for(int i8=0; i8<=9; i8++)
if(i1+i2+i3+i4==i5+i7+i6+i8)count++;
cout<<count<<endl;
}
int main()
{
int n;
cin>>n;
if(n==2)f2();
else if(n==4)f4();
else if(n==6)f6();
else if(n==8)f8();
system("pause");
return 0;
}
]]>
using namespace std;
double a[100005];
double abs(double a)
{
return a>0?a:-a;
}
int main()
{
int n,i,x=2;
double xl;
cin>>n;
for(i=1; i<=n; i++)
cin>>a[i];
xl=abs(a[2]-a[1]);
for(i=3; i<=n; i++)
{
if(abs(a[i]-a[i-1])>xl)
{
x=i;
xl=abs(a[i]-a[i-1]);
}
}
cout<<x-1<<' '<<x<<endl;
system("pause");
return 0;
}
]]>
#include<algorithm>
using namespace std;
int a[100005]={0};
int main()
{
int n,k,i;
cin>>n;
for(i=1; i<=n; i++)
cin>>a[i];
sort(a+1,a+1+n);
getchar();getchar();getchar();getchar();
cin>>k;
int temp;
for(i=1; i<=k; i++)
{ cin>>temp;cout<<a[temp]<<endl; }
system("pause");
return 0;
}
]]>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int k,i,sum=0,temp;
cin>>k;
vector<int>vec;
for(i=0; i<k; i++)
{ cin>>temp;vec.push_back(temp); }
sort(vec.begin(),vec.end());
for(i=0; i<=k/2; i++)
sum+=(vec[i]+1)/2;
cout<<sum<<endl;
system("pause");
return 0;
}
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int k,i,sum=0,temp;
cin>>k;
vector<int>vec;
for(i=0; i<k; i++)
{ cin>>temp;vec.push_back(temp); }
sort(vec.begin(),vec.end());
for(i=0; i<=k/2; i++)
sum+=(vec[i]+1)/2;
cout<<sum<<endl;
system("pause");
return 0;
}
]]>
//Ural 1022
#include<iostream>
#include<vector>
using namespace std;
int a[101][101]={0};
vector<int>vec;
int n,i,j;
bool f[101]={0};
void input()
{
cin>>n;
for(i=1; i<=n; i++)
{
j=1;
while(cin>>a[i][j],a[i][j]!=0)
j++;
}
}
void dfs(int i)
{
if(f[i])return ;
for(int j=1; a[i][j]!=0; j++)
dfs(a[i][j]);
f[i]=1; vec.push_back(i);
}
void topo()
{
for(int i=1; i<=n; i++)
if(!f[i])dfs(i);
cout<<vec[vec.size()-1];
for(int i=vec.size()-2; i>=0; i--)
cout<<' '<<vec[i];
cout<<endl;
}
int main()
{
input();
topo();
system("pause");
return 0;
}
#include<iostream>
#include<vector>
using namespace std;
int a[101][101]={0};
vector<int>vec;
int n,i,j;
bool f[101]={0};
void input()
{
cin>>n;
for(i=1; i<=n; i++)
{
j=1;
while(cin>>a[i][j],a[i][j]!=0)
j++;
}
}
void dfs(int i)
{
if(f[i])return ;
for(int j=1; a[i][j]!=0; j++)
dfs(a[i][j]);
f[i]=1; vec.push_back(i);
}
void topo()
{
for(int i=1; i<=n; i++)
if(!f[i])dfs(i);
cout<<vec[vec.size()-1];
for(int i=vec.size()-2; i>=0; i--)
cout<<' '<<vec[i];
cout<<endl;
}
int main()
{
input();
topo();
system("pause");
return 0;
}
]]>
1020. Rope
Time Limit: 1.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
Plotters have barbarously hammered N nails into an innocent plane shape, so that one can see now only heads. Moreover, pursuing their mean object, they have hammered all the nails into the vertices of a convex polygon. After that they…it is awful… have roped off the nails, so that the shape felt upset (the rope was very thin). They’ve done it as it is shown in the figure.
Your task is to find out a length of the rope.
Input
There two numbers in the first line of the standard input: N 鈥?a number of nails (1 ≤ N ≤ 100), and a real number R 鈥?a radius of heads of nails. All the heads have the same radius. Further there are N lines, each of them contains a pair of real coordinates (separated by a space) of centers of nails. An absolute value of the coordinates doesn’t exceed 100. The nails are described either in a clockwise or in a counterclockwise order starting from an arbitrary nail. Heads of different nails don’t adjoin.
Output
Output a real number with two digits precision (after a decimal point) 鈥?a length of the rope.
Sample
| input | output |
|---|---|
4 1
0.0 0.0
2.0 0.0
2.0 2.0
0.0 2.0
|
14.28
|
Problem Author: Alexander Petrov & Nikita Shamgunov
Problem Source: Ural State University Internal Contest October'2000 Junior Session
Problem Source: Ural State University Internal Contest October'2000 Junior Session
緇撴灉灝辨槸鍚勭嚎孌甸暱搴︾殑鍜屽姞涓婂崐寰勪負(fù)R鐨勫懆闀匡紝n=1鏃?闇棰濆鑰冭檻涓?br>
#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
double const pi=acos(-1.0);
int main()
{
int n=0,i=0;
double sum=0,r=0,x0,y0,x1,y1,x,y;
cin>>n>>r;
cin>>x>>y;// 淇濆瓨絎竴涓偣
x0=x1=x; y0=y1=y;
sum=2*pi*r;
for(i=1; i<n; i++)
{
cin>>x1>>y1;
sum+=sqrt( (x1-x0)*(x1-x0)+ (y1-y0)*(y1-y0) );
x0=x1; y0=y1;
}
if(n!=1) sum+=sqrt( (x1-x)*(x1-x)+ (y1-y)*(y1-y) );
printf("%.2lf\n",sum);
system("pause");
return 0;
}
#include<stdio.h>
#include<cmath>
using namespace std;
double const pi=acos(-1.0);
int main()
{
int n=0,i=0;
double sum=0,r=0,x0,y0,x1,y1,x,y;
cin>>n>>r;
cin>>x>>y;// 淇濆瓨絎竴涓偣
x0=x1=x; y0=y1=y;
sum=2*pi*r;
for(i=1; i<n; i++)
{
cin>>x1>>y1;
sum+=sqrt( (x1-x0)*(x1-x0)+ (y1-y0)*(y1-y0) );
x0=x1; y0=y1;
}
if(n!=1) sum+=sqrt( (x1-x)*(x1-x)+ (y1-y)*(y1-y) );
printf("%.2lf\n",sum);
system("pause");
return 0;
}
]]>
//Ural 1017
#include<iostream>
using namespace std;
const int MAX=505;
long long f[MAX][MAX]={0};
//f[i][j]琛ㄧず鏈鍚庝竴涓樁姊殑楂樺害涓嶈秴榪噄錛屼嬌鐢╦涓猙ricks鐨勬柟妗?nbsp; f[i][j]=f[i-1][j]+f[i-1][j-i]
int main()
{
int n,i,j;
cin>>n;
f[0][0]=1;
for(i=1; i<=n; i++)
{
for(j=n; j>=i; j--)
f[i][j]=f[i-1][j]+f[i-1][j-i];
for(j=0; j<=i-1; j++)
f[i][j]=f[i-1][j];
}
cout<<f[n][n]-1<<endl;
//system("pause");
}
]]>
#include<iostream>
using namespace std;
int a[10]={0};
int main()
{
int n,i;
cin>>n;
if(n==0){cout<<10<<endl;return 0;}
else if(n==1){cout<<1<<endl; return 0;}
int p=n;
for(i=9; i>=2; )
{
if(p%i==0){a[i]++; p/=i;}
else i--;
}
if(p!=1){ cout<<-1<<endl; return 0; }
for(i=2; i<=9; i++)
for(int j=1; j<=a[i]; j++)
cout<<i;
cout<<endl;
system("pause");
return 0;
}
using namespace std;
int a[10]={0};
int main()
{
int n,i;
cin>>n;
if(n==0){cout<<10<<endl;return 0;}
else if(n==1){cout<<1<<endl; return 0;}
int p=n;
for(i=9; i>=2; )
{
if(p%i==0){a[i]++; p/=i;}
else i--;
}
if(p!=1){ cout<<-1<<endl; return 0; }
for(i=2; i<=9; i++)
for(int j=1; j<=a[i]; j++)
cout<<i;
cout<<endl;
system("pause");
return 0;
}
]]>
| Accepted | 0.062 | 1 241 KB |
#include<iostream>
using namespace std;
int p1[100000]={0},p2[100000]={0};
int ne1[100000]={0},ne2[100000]={0};
int main()
{
int n1,n2,i,t;
cin>>n1;
for(i=0; i<n1; i++)
{
cin>>t;
if(t>=0)p1[t]=1;
else ne1[-t]=1;
}
cin>>n2;
for(i=0; i<n2; i++)
{
cin>>t;
if(t>=0)p2[t]=1;
else ne2[-t]=1;
}
bool flag=0;
for(i=0; i<=100000; i++)
{
if(p1[i])
if(i<=10000)
{ if(p2[10000-i]){flag=1;break;}}
else if(ne2[i-10000]){flag=1;break; }
}
if(flag==0)
for(i=0; i<=100000; i++)
{
if(p2[i])
if(i<=10000)
{ if(p1[10000-i]){flag=1;break;} }
else if(ne1[i-10000]){flag=1;break; }
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
// system("pause");
return 0;
}
using namespace std;
int p1[100000]={0},p2[100000]={0};
int ne1[100000]={0},ne2[100000]={0};
int main()
{
int n1,n2,i,t;
cin>>n1;
for(i=0; i<n1; i++)
{
cin>>t;
if(t>=0)p1[t]=1;
else ne1[-t]=1;
}
cin>>n2;
for(i=0; i<n2; i++)
{
cin>>t;
if(t>=0)p2[t]=1;
else ne2[-t]=1;
}
bool flag=0;
for(i=0; i<=100000; i++)
{
if(p1[i])
if(i<=10000)
{ if(p2[10000-i]){flag=1;break;}}
else if(ne2[i-10000]){flag=1;break; }
}
if(flag==0)
for(i=0; i<=100000; i++)
{
if(p2[i])
if(i<=10000)
{ if(p1[10000-i]){flag=1;break;} }
else if(ne1[i-10000]){flag=1;break; }
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
// system("pause");
return 0;
}
]]>
| Accepted | 0.25 | 1 237 KB |
#include<iostream>
#include<cmath>
using namespace std;
double a[300000];
int main()
{
long long i,j=0;
while(cin>>i)
{
a[j++]=sqrt(double(i));
}
for(i=j-1;i>=0; i-- )
printf("%.4lf\n",a[i]);
system("pause");
return 0;
}
]]>
| Accepted | 0.031 | 201 KB |
#include<iostream>
#include<cmath>
using namespace std;
int x[200],y[200];
double ex=1e-12;
int main()
{
int n,i,cnt,max=0,j;
cin>>n;
for(i=0;i<n; i++)
cin>>x[i]>>y[i];
for(i=0; i<n; i++)
for(j=i+1; j<n; j++)
{
cnt=2;
int a=y[j]-y[i],
b=-(x[j]-x[i]),
c=-(y[j]-y[i])*x[i] + y[i]*(x[j]-x[i]);
for(int k=0; k<n; k++)
if(k==j||k==i)continue;
else if(a*x[k]+b*y[k]+c==0)cnt++;
if(cnt>max)max=cnt;
}
cout<<max<<endl;
system("pause");
return 0;
}
]]>
| //Accepted | 0.062 | 277 KB |
1 #include<iostream>
2 #include<cmath>
3 using namespace std;
4 const int cnt=15000;
5 int a[cnt+1]={0};
6 bool isprime(int n)
7 {
8 if(n==2||n==3||n==5||n==7)return 1;
9 if(n%2==0||n<2)return 0;
10 double t=sqrt(n*1.0);
11 for(int i=3; i<=t; i+=2)
12 if(n%i==0)return 0;
13 return 1;
14 }
15
16 int main()
17 {
18 int n,i,k,count=0;
19 cin>>n;
20 for(i=2; count<=cnt; i++)
21 {
22 if(isprime(i))a[++count]=i;
23 }
24 for(i=1; i<=n; i++)
25 { cin>>k; cout<<a[k]<<endl;}
26 system("pause");
27 return 0;
28 }
29
]]>
/*
| Accepted | 0.031 | 233 KB |
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
double p2=sqrt(2.0),
pi=acos(-1.0),
tri,//瑙掑害
area=0;
int l,r;
cout.precision(3);
cin>>l>>r;
if(r<l*1.0/2)cout<<fixed<<pi*r*r<<endl;
else if(r>l*1.0*p2/2)cout<<fixed<<(double)l*l<<endl;
else {
tri=acos(l*1.0/(2*r));
area+=(l*1.0*r*sin(tri)/2);
area+=(pi/2-2*tri)*r*r/2;
area*=4;
cout<<fixed<<area<<endl;
}
system("pause");
return 0;
}
]]>
#include<iostream>
//#include<limit.h>
#include<cmath>
using namespace std;
int w[21]={0};
int n,total=0,ans=INT_MAX;
void search(int i,int now)
{
if(i>n)
{
if(ans>abs(total-2*now))ans=abs(total-2*now);
return ;
}
search(i+1,now+w[i]);
search(i+1,now);
}
int main()
{
cin>>n;
for(int i=1;i<=n; i++)
{ cin>>w[i];total+=w[i];}
search(1,0);
cout<<ans<<endl;
return 0;
}
DP錛?1鑳屽寘 寰瀹歸噺涓簍otal/2鐨勫寘閲岃鐭沖ご錛屼環(huán)鍊間笌閲嶉噺鐩哥瓑
//
#include<iostream>
using namespace std;
const int SIZE=21;
int w[SIZE]={0};
int dp[SIZE*100000/2+5]={0};
int main()
{
int n,i,total,v,j;
cin>>n;
for(i=1,total=0; i<=n ; i++ )
{ cin>>w[i]; total+=w[i]; }
v=total/2;
for(i=1; i<=n; i++)
for(j=v; j>=w[i]; j-- )
if(dp[j]<dp[j-w[i]]+w[i])
dp[j]=dp[j-w[i]]+w[i];
cout<<total-2*dp[v]<<endl;
return 0;
}
//
| Accepted | 0.031 | 3933 KB |
#include<iostream>
using namespace std;
const int SIZE=21;
int w[SIZE]={0};
int dp[SIZE*100000/2+5]={0};
int main()
{
int n,i,total,v,j;
cin>>n;
for(i=1,total=0; i<=n ; i++ )
{ cin>>w[i]; total+=w[i]; }
v=total/2;
for(i=1; i<=n; i++)
for(j=v; j>=w[i]; j-- )
if(dp[j]<dp[j-w[i]]+w[i])
dp[j]=dp[j-w[i]]+w[i];
cout<<total-2*dp[v]<<endl;
return 0;
}
]]>