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joj 1894 高精度除法有巧解

luis — Sat, 18 Jul 2009 09:38:00 GMT

Given any integer 0 <= n <= 10000 not divisible by 2 or 5, some multiple of n is a number which in decimal notation is a sequence of 1's. How many digits are in the smallest such a multiple of n?

Sample input

3
7
9901

Output for sample input

3
6
12

可以不用高精度除法，直接如下�Q�很巧妙�Q�不通用�?br>

#include 
int main()
{
int n, a, b;
while (EOF != scanf("%d", &n))
{
a = 1;
b = 1;
while (a)
{
a = (a * 10 + 1) % n;
++b;
}
printf("%d\n", b);
}
}

高精度除�?br>----
#include
#include
using namespace std;
int a[9]={1,11,111,1111,11111,111111,1111111,11111111,111111111};
const int MOD=11111;
bool test(int i,int j)
{
  if(j<10)
  {
   if(a[j-1]%i==0)
   return true;
   return false;
  }
  else
  {
   int temp=0;
   int k=j%5;
   int p=j/5;
   if(k==0)
   {
    for(int m=1;m<=p;m++)
      {
     temp*=100000;
        temp+=MOD;
        temp%=i;

       }
       if(temp==0)
       return true;
       else
       return false;
     }
     else
     {
    int m,f=1;
    for(m=1;m<=p;m++)
      {
        temp*=100000;//�׃��5�?位除�Q�应当乘�?0^5;
     temp+=MOD;
        temp%=i;

       }
       for(m=0;m       f*=10;
     if((temp*f+a[k-1])%i==0)
     return true;
     else
     return false;
     }

  }
}
int main()
{
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
int i;
while(cin>>i)
{
  if(i==1)
  {
   cout<<'1'<   continue;
  }
  for(int j=2;;j++)
  {
   if(test(i,j))
   {
    cout<    break;
   }
      }
}

//system("PAUSE");
return 0;
}

luis 2009-07-18 17:38 发表评论

joj 2391 words 深搜剪枝

luis — Sun, 05 Jul 2009 04:33:00 GMT

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	439	74	Standard

Io and Ao are playing a word game. They say a word in the dictionary by turns. The word in the dictionary only contains lowercase letters. And the end character of the former said word should be the same as the start character of the current said word. They can start the game from any word in the dictionary. Any word shouldn't be said twice. Now, we define the complexity of the game that is the sum length of all words said in the game. Give you a dictionary, can you tell me the max complexity of this word game?

Input

The first line contains a single positive integer n(0 < n <=12). Next n lines are n words in the dictionary. The length of each word will not exceed 100.

Output

A single integer represents the complexity of the game.

Sample Input

3
joj
jlu
acm
6
cctv
redcode
lindong
we
love
programming
3
daoyuanlee
come
on

Sample Output

6
11
10

Problem Source: provided by loon

#include
#include
using namespace std;
struct S
{
string a;
char begin;
char end;
int length;
}s[13];
int visited[13];
int temp;
void search(int a,int num,int pre)
{

  for(int i=0;i  {
   if(s[i].begin==s[pre].end&&visited[i]==0)
   {
    visited[i]=1;
    search(a+s[i].length,num,i);
    if(a+s[i].length>temp)temp=a+s[i].length;
    visited[i]=0;
   }
  }
}
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
   int num;
   while(cin>>num)
   {
    int i;
  temp=0;
  for( i=0;i  {
   cin>>s[i].a;
   s[i].length=(s[i].a).size();
   s[i].begin=(s[i].a)[0];
   s[i].end=(s[i].a)[s[i].length-1];
   if(s[i].length>temp)
   temp=s[i].length;
     }
  for(i=0;i     {
   memset(visited,0,sizeof(visited));
   visited[i]=1;
   search(s[i].length,num,i);
  }
  cout< }

//system("PAUSE");
return 0;
}

以上代码��时。完全可以剪枝�?br>举个例子
abc
cbd
dbm
dbacmdp
我的�E�序一直搜啊搜�Q�每�ơ搜完都重新开始。比如在以a开头后�Q�搜到c�Q�下�ơ再搜烦(ch��)时直接利用c的结果，�q�是深搜的特点决定的�Q�！�Q?br>*************************
�q�种�c�M��的有序搜索都可以�?nbsp;    * 备忘录方�?
**************************
#include
#include
using namespace std;
int num;
struct S
{
string a;
char begin;
char end;
int length;
}s[13];
int visited[13];
int temp;
int sum[13];
int search(int pre)//·µ»Ø´ÓpreµãÒÔºóµÄ×ÜµÄÖµ
{
  int j=s[pre].length,k=0;
  for(int i=0;i  {
   if(s[i].begin==s[pre].end&&visited[i]==0&&i!=pre)//必须要有I�Q?pre
   {
    visited[i]=1;
    k=search(i)+s[pre].length;
    if(k>j)j=k;
   }
   else
   {
    if(s[i].begin==s[pre].end&&i!=pre)//必须要有i!=pre
    return sum[i]+s[pre].length;//相当于备忘录�Q�而且无需visited[i]=0;
   }
  }
  sum[pre]=j;
  return j;
}
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
   while(cin>>num)
   {
    int i,j;
  temp=0;
  j=0;
  for( i=0;i  {
   cin>>s[i].a;
   s[i].length=(s[i].a).size();
   s[i].begin=(s[i].a)[0];
   s[i].end=(s[i].a)[s[i].length-1];
   if(s[i].length>temp)
   temp=s[i].length;
     }
  for(i=0;i     {
   memset(visited,0,sizeof(visited));
            memset(sum,0,sizeof(sum));
   visited[i]=1;
   j=search(i);
   if(j>temp)
   temp=j;
  }
  cout< }

//system("PAUSE");
return 0;
}

因�ؓ(f��)I�Q?pre又错了几下�?br>以后debug��量自己用眼睛看�Q�更省时��_(d��)��Q�！�Q�！�Q�！�Q�！

luis 2009-07-05 12:33 发表评论

luis — Sat, 04 Jul 2009 01:27:00 GMT

Energy

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	10240K	717	196	Standard

Mr. Jojer is a very famous chemist. He is doing a research about behavior of a group of atoms. Atoms may have different energy and energy can be positive or negative or zero, e.g. 18 or -9. Absolute value of energy can not be more than 100. Any number of continuous atoms can form an atom-group. Energy of an atom-group is defined by the sum of energy of all the atoms in the group. All the atoms form an atom-community which is a line formed by all the atoms one by one. Energy of an atom-community is defined by the greatest energy of an atom-group that can be formed by atoms in the atom-community. The problem is, given an atom-community, to calculate its energy.

Input

The input contains several test cases. Each test case consists of two lines describing an atom-community. The first line of each test case contains an integer N(N<=1000000), the number of atoms in the atom-community. The second line of each test case contains N integers, separated by spaces, each representing energy of an atom, given in the order according to the atom-community. The last test case marks by N=-1, which you should not proceed.

Output

For each test case(atom-community description), print a single line containing the energy.

Sample Input

5
8 0 6 4 -1
-1

Sample Output

理解题意很重要，题目的意思是�?在m个中�?nbsp; �q�箋的n个atom能量�?使其最大�?br>�E�序中sumtemp�Q�sum。sumtemp��定的是其左边界�Q�sum��定其右边界�?br>sumtemp��定左边前n个数之和��的最大的n�Q�且�W�n个数昄��Q�然后从n+1开始选数�?br>sum��定了右边界使其最大�?br>求和最大都可以用这�U�思�\�Q�！�Q�！�Q�！�Q�！�Q�！
举例
1�Q?�Q?4�Q?�Q?�Q?2�Q?�Q?�Q?6�Q?�Q?3�Q?�Q?0
��L(f��ng)��sumt1=1,sum=1
sumt2=2;sum=1+2=3;
sum3=sum2-4=-1;则sumtemp=0;sum不变�?br>sum4=sumtemp+4;sumsum5=sumtemp+2�Q�sum��M��sum只有在sum一�l�dp.
#include"stdio.h"
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int n;
int a;
while(scanf("%ld",&n),n!=-1)
{
      long long sum=-0x7fffffff,sumtemp=-0x7fffffff;
   for(long i=0;i   {
    scanf("%d",&a);
    if(sumtemp>0)
   sumtemp+=a;
    else
   sumtemp=a;
    if(sumtemp>sum)
   sum=sumtemp;

}
printf("%lld\n",sum);

}
return 0;
}

luis 2009-07-04 09:27 发表评论

luis — Fri, 03 Jul 2009 15:29:00 GMT

错误代码�Q�过不了的数�?4�Q?5 10 17 16 12 1 10 20 17 19 4 5 9 5 可能解：(x��)(20 15 5 ),(19 17 4 ),( 17 12 10 1 ), ( 16 10 9 5 )
下面的代�?#8220;no”�Q�因�?20 19 1),(17接下来凑不出�?nbsp;)
#include
#include
#include
using namespace std;
int length[21];
int mark[21];
bool cmp(int x,int y)
{
return x>y;
}

int dfs(int sum,int flag,int n,int time)//´Ótime¿ªÊ¼ËÑË÷
{
  if(sum==0)return 1;
    if(time  {
   for(int k=time;k   {
    if(mark[k]<0&&sum-length[k]>=0)
    {
     mark[k]=flag;
     if(dfs(sum-length[k],flag,n,k+1))
     return 1;
     else
     mark[k]=-1;
       }
   }
  }

return 0;
}
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int num,n;

cin>>num;
while(num--)
{
  int sum=0;
  cin>>n;
  for(int k=0;k  {
   cin>>length[k];
   mark[k]=-1;
   sum+=length[k];
  }
  sort(length,length+n,cmp);
  if(sum%4!=0||length[0]>sum/4)
  cout<<"no"<  else
  {
   sum/=4;
   if(dfs(sum,1,n,0))
    {
   if(dfs(sum,2,n,0))
         {
     if(dfs(sum,3,n,0))
     cout<<"yes"<     else
     cout<<"no"<        }
        else
        cout<<"no"<          }
         else
         cout<<"no"<  }
    }
//system("PAUSE");
return   0;
}
正确的解法应当是�Q�在一��h��索�?br>下面是一�D�超时的代码
#include
#include
#include
using namespace std;
int length[21];
int mark[21];
int flag=0;
int target;
bool cmp(int x,int y)
{
    return x>y;
}

int func(int i)
{
  while(mark[i]>0)
  i++;
  return i;
}
void dfs(int sum,int n,int time,int level)//从序号�ؓ(f��)time的开始搜�?
{
  if(flag)return ;
  if(sum==0&&level==3)
  {
  flag=1;
  return;
     }
     if(sum==0)
     {
   level++;
   dfs(target,n,func(0),level);
  }
    else if(time  {
   for(int k=time;k   {
    if(k>1)
               {

                if(length[k]==length[k-1]&&!mark[k-1])
                    continue;//昄��
               }
                if(sum    if(mark[k]<0&&sum-length[k]>=0)
    {
     mark[k]=1;
     dfs(sum-length[k],n,func(k+1),level);
     mark[k]=-1;
       }
   }
  }
}
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int num,n;

cin>>num;
while(num--)
{
  int sum=0;
  cin>>n;
  for(int k=0;k  {
   cin>>length[k];
   mark[k]=-1;
   sum+=length[k];
  }
  sort(length,length+n,cmp);
  if(sum%4!=0||length[0]>sum/4)
  cout<<"no"<  else
  {
   sum/=4;
   target=sum;
   dfs(sum,n,0,0);
       if(flag==0)
       cout<<"no"<       else
       cout<<"yes"<  }
    }
//system("PAUSE");
return   0;
}

上面��时��p��时原因是减枝不彻底。level标记也不太好
军_��另外炉灶,先看看别人的代码
#include
#include
using namespace std;
int s[21];
int v[21];
int len;
int m;
int dfs(int cur,int num,int beg,int fin)
{
int solve(int );
if(num==1)
  return 1;
if(cur==len)
{
return solve(num-1);
}
   for(int i=beg;i>=fin;i--)
   {
    if(!v[i]&&cur+s[i]<=len)
    {
     v[i]=1;
     if(dfs(cur+s[i],num,i-1,fin))
      return 1;
     v[i]=0;
    }
   }
   return 0;
}

int solve(int edge_num)
{
     int i;
for(i=m;i>=1;i--)
   if(!v[i])
   {
    v[i]=1;
    if(dfs( s[i],edge_num,i-1,1))
     return 1;
    v[i]=0;
   }
return 0;
}

int main()
{
int n;
cin>>n;
while(n--)
{

  cin>>m;
  int i;
  int sum=0;
  for(i=1;i<=m;i++)
  {
   cin>>s[i];
   sum+=s[i];
   v[i]=0;
  }

  len=sum/4;
  if(4*len!=sum)
  {
   cout<<"no"<   continue;
  }
  sort(s+1,s+m+1);
  if(s[m]>len)
  {
   cout<<"no"<   continue;
  }
  if(solve(4))
   cout<<"yes"<  else
   cout<<"no"< }
return 0;
}//剪枝�q�可以做得更好！�Q�！
人家的solve写得好！�Q�，�l�合dfs.

luis 2009-07-03 23:29 发表评论

luis — Thu, 02 Jul 2009 08:05:00 GMT

Tug of War

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	15s	8192K	479	114	Standard

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

The input may contain several test cases.

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

Output for Sample Input

190 200

利用dp思想 �Q�n为偶数时求出s(n,n/2)�Q�n为奇数时也是s(2n,n/2)�Q�和sum/2最接近的那个。非常经典的思�\�?br>S(k, 1) = {A[i] | 1<= i <= k}
S(k, k) = {A[1]+A[2]+…+A[k]}
S(k, i) = S(k-1, i) U {A[k] + x | x属于S(k-1, i-1) }
//一下代码只能用于sum特别��的情况�Q�否则会(x��)��时�Q�！�Q�！�Q�！�Q�！�Q�！�Q?br>#include
#include
#define MAX 101
#define min(a,b) ((a)<(b) ? (a) : (b))
using namespace std;

int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int num;
int a[MAX],i,j,k,l,m,NUM;
bool s[MAX][2500];
while(cin>>num)
{
  int sum=0;
    for(i=1;i<=num;i++)
   {
cin>>a[i];
sum+=a[i];
   }
   if(num%2==0)NUM=num/2;
   else NUM=num/2+1;

    for(i=0;i<=num;i++)
     for(j=0;j<=sum/2;j++)
      s[i][j]=false;//表示取i个物品能否达到重量是j.

   s[0][0]=true;
   for(k=1;k<=num;k++)//必须在最外层�Q�元素不能重�?
   for(j=min(k,NUM);j>=1; j--)//递减的结果是使得不会(x��)出现在同一层次的互为因�?、、、、、、、、、、、��y妙的实现了课本上的序偶对生成法�?br>   for(i=a[k];i<=sum/2;i++)
   {
  if(s[j-1][i-a[k]])
  s[j][i]=true;
}

for(i=sum/2; i>=0; i--) {
    if(s[NUM][i]) {
        cout <        break;
    }
}

}
//system("PAUSE");
return 0;
}
下一�ơ实��C��个序偶生成法�?/p>

#include
#include
using namespace std;

int a[101];
bool b[101][45002];

int main(){
// freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
    int N,M,i,j,k;
    while(scanf("%d",&N)!=EOF){
         memset(b,0,sizeof(b));
         a[0]=M=0;
        for(i=1;i<=N;i++){
             scanf("%d",a+i);
             M+=a[i];
         }
         b[0][0]=1;
        for(k=1;k<=N;k++){
            for(i=1;i<=N/2;i++){
                for(j=M/2;j>=0;j--){
                    if(b[i-1][j]){
                         b[i][j+a[k]]=1;
                     }
                 }
             }
         }
        for(i=M/2,j=M/2+1;i>=0;i--,j++){
            if(b[N/2][i]){
                 printf("%d %d\n",i,M-i);
                break;
             }
            if(b[N/2][j]){
                 printf("%d %d\n",M-j,j);
                break;
             }
         }
     }
    return 0;
}

luis 2009-07-02 16:05 发表评论

joj 2250 my friends深度优先搜烦(ch��)讲解

luis — Mon, 29 Jun 2009 08:13:00 GMT

I have my friends to visit. For some reason, I can only visit some of my friends. So I want see my friends as many as possible. Thus I must choose the longest way. Your goal is to help me developing a program that computes the length of the longest path that can be constructed in a given graph from a given starting point (My residence). You can assume that the graph has no cycles (there is no path from any node to itself), so I will reach my destination in a finite time. In the same line of reasoning, nodes are not considered directly connected to themselves.

Input

The input consists of a number of cases. The first line on each case contains a positive number n (1 < n <= 100) that specifies the number of points in the graph. A value of n = 0 indicates the end of the input. After this, a second number s is provided, indicating the starting point in my journey (1 <= s <= n). Then, you are given a list of pairs of places p and q, one pair per line. The pair "p q" indicates that I can visit q after p. A pair of zeros ("0 0") indicates the end of the case. As mentioned before, you can assume that the graphs provided will not be cyclic.

Output

For each test case you have to find the length of the longest path that begins at the starting place. You also have to print the number of the final place of such longest path. If there are several paths of maximum length, print the final place with smallest number.

Print a new line after each test case.

Sample Input

Sample Output

Case 1: The longest path from 1 has length 1, finishing at 2.
Case 2: The longest path from 3 has length 4, finishing at 5.
Case 3: The longest path from 5 has length 2, finishing at 1.

提交�?0�ơ，AC 3�ơ，���时4�ơ，wa 3�ơ�?/pre>
很无语�?/pre>
应该是动态规划最好，但是我不是很熟，用了搜烦(ch��)。以下是���时的代�?/pre>
#include
#include
using namespace std;
int path[101][101];
int mark[101];
int len[101];//
int end[101];
int dfs(int start,int num)//�q�回从当前点出发的最大长�?
{
if(mark[start]==1)return len[start];
mark[start]=1;
end[start]=start;
for(int i=1;i<=num;i++)
{
if(path[start][i])
{
if(len[start]end[i])
end[start]=end[i];
}
}
mark[start]=0;//�q�句坚决不需�?
return len[start];
}
int main()
{
// freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
int j,k,turn=0;
int start,num;
while(cin>>num)
{
turn++;
if(num==0)break;
memset(path,0,sizeof(path));
memset(mark,0,sizeof(mark));
memset(len,0,sizeof(len));
memset(end,0,sizeof(end));
cin>>start;
while(cin>>j>>k)
{
if(j==0)break;
path[j][k]=1;
}
cout<<"Case "<
以下是ac的代�?/pre>
#include
#include
using namespace std;
int path[102][102];
int mark[102], len[102],end[102];
int dfs(int start,int num)//�q�回从当前点出发的最大长�?
{
if(mark[start]==1)return len[start];
mark[start]=1;
end[start]=start;
len[start]=0;
int i,t;
for( i=1;i<=num;i++)
{
if(path[start][i])
{
t=dfs(i,num)+1;
if(t>len[start])
{
len[start]=t;
end[start]=end[i];
}
else
if(len[start]==t)
{
if(end[start]>end[i])
end[start]=end[i];
}
}
}
return len[start];
}
int main()
{
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
int j,k,turn=0;
int start,num;
while(cin>>num,num)
{
turn++;
memset(path,0,sizeof(path));
memset(mark,0,sizeof(mark));
memset(len,0,sizeof(len));
memset(end,0,sizeof(end));
cin>>start;
while(cin>>j>>k,j||k)
{
path[j][k]=1;
}
dfs(start,num);
cout<<"Case "<
不妨执行一�?/pre>
5
3
1 2
3 5
3 1
2 4
4 5
0 0
先是len[3]=0;end[3]=3;flag[3]=1;
再执行t=dfs(1)+1,
转入dfs�Q?�Q�；len[1]=0;end[1]=1;flag[1]=1;
再执�?span style="COLOR: #0000ff">t=dfs(2)+1;
转入dfs(2),len[2]=0;end[2]=2;flag[2]=1;
再执行t=dfs(4)+1
转入dfs(4),len[4]=0;end[4]=4;flag[4]=1;
再�{�?span style="COLOR: red">t=dfs�Q?)+1;
转入dfs(5),len[5]=0;end[5]=5;flag[5]=1;return(len[5]=0);
则t=1;t>len[4];len[4]=1;end[4]=end[5]=5;再看4没了其他盔R��元素。dfs(4)=return(len[4])=1;
t=dfs(4)+1=2;len[2]=t=2;end[2]=end[4]=5;再看2没了其他盔R��元素�Q�dfs(2)=return(len(2)=2;
再看t=dfs(2)+1=3;len[1]=t=3;end[1]=en[2]=5;再看1有没有其他相��d��素，dfs(1)=return(len(1)=3
再执行t=dfs(1)+1,len[3]=4;end[3]=end[1]=5;再看3有没有其他相��d��素，有dfs(5),已经遍历��C���Q�所以dfs�Q?�Q�return len�?】�?br>没有影响�?br>假设改�ؓ(f��)
5
3
5 2
3 5
3 1
2 4
4 1
0 0
执行时会(x��)�?->1>�q�时�?�l�点len[1]已经求的   3>5>2>4>1len[1]已知�?/span>




luis 2009-06-29 16:13 发表评论

luis — Mon, 29 Jun 2009 03:44:00 GMT

启发1�Q�double�q�算速度比Int�?br>启发2�Q�m选N的组合数�Q�如果n已知�Q�只需循环卛_��
#include
#include
using namespace std;
int main()
{
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
double n[1001];
int i,j,k,m;
int flag,num;
while(cin>>num,num)
{
  flag=0;
  for(m=1;m<=num;m++)
  {
   cin>>n[m];
  }
  for(i=1;i<=num;i++)
  {
    for(j=1;j<=num;j++)
     {
       for(k=1;k<=num;k++)
       {
      if(n[i]+n[j]+n[k]==0)
     {
    flag=1;break;
      }
       }
       if(flag==1)break;
   }
   if(flag==1)break;
   }
  if(flag==1)cout<<"yes"<  else cout<<"no"< }

//system("PAUSE");
return 0;
}

luis 2009-06-29 11:44 发表评论

luis — Sun, 28 Jun 2009 14:26:00 GMT


Your teacher has given you some problems to solve. You must first solve problem 0. After solving each problem i, you must either move on to problem i+1 or skip ahead to problem i+2. You are not allowed to skip more than one problem. For example, {0, 2, 3, 5} is a valid order, but {0, 2, 4, 7} is not because the skip from 4 to 7 is too long.
You are given N integers, where the ith integer indicates how much you like problem i. The teacher will let you stop solving problems once the range of pleasantness you've encountered reaches a certain threshold. Specifically, you may stop once the difference between the maximum and minimum pleasantness of the problems you've solved is greater than or equal to a integer K. If this never happens, you must solve all the problems.
Input
The input contains several test cases. For each case it contains two positive integer N, K (1<=N<=50, 1<=K<=1000), then N integers follow in the next line, the ith integer ( >=1 and <= 1000 ) indicates how much you like problem i.

Output
For each test case, output the minimum number of problems you must solve to satisfy the teacher's requirements.
Sample Input
3 2
1 2 3
5 4
1 2 3 4 5
Sample Output
2
3 

很笨但是很实用！
#include
int a[51];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
int i,j;
for(i=0;i=k||a[j]-a[i]>=k)
{
flag=1;
int step;
if(j==0)
{
step=(i+1)/2+1;
if(min>step)
min=step;
}
else
{
step=(i-j+1)/2+(j+1)/2+1;
if(min>step)
min=step;
}
}
}
if(flag)
printf("%d\n",min);
else
printf("%d\n",n);
}
return 0;
}

luis 2009-06-28 22:26 发表评论

计算n�Q�的位数�Q�非常��y妙啊�Q�！

luis — Sun, 28 Jun 2009 09:22:00 GMT

#include
#include
using namespace std;
int main()
{
int n;
double i;
long  double sum;
int sum2;
while(cin>>n,n)
{
sum=1.0;
for(i=1.0;i<=n;i++)
sum+=log10(i);
sum2=sum;
printf("%d\n",sum2);
}
return 0;
}

���乘法�{化�ؓ(f��)log加法�Q�！�Q?/pre>


luis 2009-06-28 17:22 发表评论

luis — Sat, 27 Jun 2009 02:44:00 GMT

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.

Output

Numbers Q, one on each line.

Sample Input

3
5
0

Sample Output

1
2

�Ҏ(gu��)��1�Q�动态规�?br>
#include
#include
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
double f[501][501]={0};
double s;
int i,j,k,n;
for(i=3;i<=500;i++)
for(j=1;j<=(i-1)/2;j++)
    f[i][j]=1;
for(i=3;i<=500;i++)
    for(j=1;j<=(i-1)/2;j++)
    {
for(k=j+1;k<=(i-j-1)/2;k++)
   f[i][j]=f[i-j][k];
}
    while(scanf("%d",&n),n) {
        s=0;
        for(i=1;i<=(n-1)/2;i++) s+=f[n][i]; //f[n]=f[n][1]+f[n][2]+-----+f[n][floor((i-1)/2)]
        printf("%.0f\n",s);
    }
//system("PAUSE");
return   0;
}
更妙的方法：(x��)生成函数�?br>计算(1+x)(1+x^2)(1+x^3)-----,x^n的系数即为所�?br>int i,j;
double ans[510]={1,1};//已经把ans[1]和ans[0]赋�ؓ(f��)1了，其余�?
for(i=2;i<=500;i++) {
        for(j=500;j>=0;j--) {
            if(i+j<=500) ans[i+j]+=ans[j];
        }
    }

先计��?1+x)(1+x^2)
再计��?1+x)(1+x^2)   *(1+x^3)

luis 2009-06-27 10:44 发表评论

map�l�典用法+多个�I�格隔开的case 获取行字�W�串

luis — Sat, 27 Jun 2009 01:05:00 GMT

#include<iostream>
#include<string>
#include<map>
using namespace std;

int main()
{
    //freopen("s.txt","r",stdin);
  //freopen("key.txt","w",stdout);
    string s;
    int g=0;
    while(getline(cin,s))
    {
        if (g) cout<<endl;
        g=1;
        int sum=0;
        map<string,int> a;
        while(s!="")
        {
            a[s]++;
            sum++;
            getline(cin,s);
        }
        cout.flags(ios::fixed);
        cout.precision(4);
        for (map<string,int>::iterator p=a.begin();p!=a.end();p++)
            cout<<p->first<<" "<<100.0*p->second/sum<<endl;
    }
    return 0;
}

#include
#include
#include
using namespace std;

int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
    string s;
    int g=0;
    while(getline(cin,s))
    {
        if (g) cout<        g=1;
        int sum=0;
        map a;
        while(s!="")
        {
            a[s]++;
            sum++;
            getline(cin,s);
        }
        cout.flags(ios::fixed);
        cout.precision(4);
        for (map::iterator p=a.begin();p!=a.end();p++)
            cout<first<<" "<<100.0*p->second/sum<    }
    return 0;
}

luis 2009-06-27 09:05 发表评论

luis — Thu, 14 May 2009 08:32:00 GMT

Sum It Up

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	424	182	Standard

Given a specifie d total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then there are four different sums that equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

Input

The input file will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers . If n = 0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.

Output

For each test case, first output a line containing `Sums of ', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line `NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.

Sample Input

4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0

Sample Output

Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25

#include<iostream>
#include<cstdlib>
#include<memory>
using namespace std;
int mount;
int l;
int a[12][2];
int mark;
  void output()
  {
    mark++;
    int temp[12];
    int flag=0;
    for(int i=0;i<mount;i++)
    {
     if(a[i][1])
    {
      if(flag==0)
      flag++;
      else
      cout<<"+";
     cout<<a[i][0];
     }
    }
     cout<<endl;
  }
  void dfs(int sum,int nowv,int pre)
  {
       int value;
       if(nowv==sum)
       {
          output();
       }
       else
       {
           for(int j=pre+1;j<mount;j++)
           {
             value=a[j][0];
             if(nowv+value<=sum&&value!=l)
             {
             a[j][1]=1;
             dfs(sum,nowv+value,j);
             a[j][1]=0;
             }
           }
       }
       if(pre!=-1)
       l=a[pre][0];
  }
  int main()
  {
  //freopen("s.txt","r",stdin);
  //freopen("key.txt","w",stdout);
  int sum,i,j;
  cin>>sum>>mount;
  while(sum!=0)
  {
    for(i=0;i<mount;i++)
    {
       cin>>a[i][0];
       a[i][1]=0;
    }

    mark=0;
    l=0;
    cout<<"Sums of "<<sum<<":"<<endl;
    dfs(sum,0,-1);
    if(mark==0)
    cout<<"NONE"<<endl;
    cin>>sum>>mount;
  }
  //system("PAUSE");
  return 0;
  }

隄��Q?�Q�有序递减�Q�这个好处理�Q�只需在向前推时dfs(i)for(int j=i+1...)卛_��
2�Q�去掉无效的重复搜烦(ch��)�Q�这个烦(ch��)一�?比如和�ؓ(f��)11�Ӟ��序列�?,6,6,2,2,2,1�Q?+2+2+1昄��是可以的�Q�但�?+6不可以�?+6+6更不可以
我们可以�q�样惻I��如果��着往前推是，如果不超�q�范��_(d��)��比如6+2+取后一个数�Ӟ��2可以取，但是6+6�Q�超�q?1�Q�要回溯的时候，记录下此时的6�Q�用下面的代�?br>
if(pre!=-1)
l=a[pre][0];

在比较if(nowv+value<=sum&&value!=l)此时不符合，下一�?取不��C��Q�避免了重复�?/div>

luis 2009-05-14 16:32 发表评论