Ads Proposal

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1063    Accepted Submission(s): 410

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<string.h>
using namespace std;
const int MAXN=100005;
const int MAXM=500005;
int U[MAXM],C[MAXM],L[MAXM];
int rank[MAXM];
int hash[MAXN];
long long res[MAXM];
int cmp(const void *a,const void *b)
{
    return C[*(int *)b]-C[*(int *)a];
}    
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int T;
    int iCase;
    int i;
    iCase=0;
    scanf("%d",&T);
    int N,M,Q;
    int k;
    while(T--)
    {
        iCase++;
        scanf("%d%d%d",&N,&M,&Q);
        for(i=0;i<M;i++)
        {
            scanf("%d%d%d",&U[i],&C[i],&L[i]);
            rank[i]=i;
        }  
        memset(hash,0,sizeof(hash));
        memset(res,0,sizeof(res));  
        qsort(rank,M,sizeof(rank[0]),cmp);
        for(i=0;i<M;i++)
        {
            res[++hash[U[rank[i]]]]+=L[rank[i]];
        }   
        for(i=1;i<=M;i++)
           res[i]+=res[i-1];
        printf("Case #%d:\n",iCase);
        while(Q--)
        {
            scanf("%d",&k);
            if(k>=M) printf("%I64d\n",res[M]);
            else printf("%I64d\n",res[k]);
        }     
    }    
    return 0;
}    

Hexadecimal is very important and useful for computer programmers. You are requested to provide a hexadecimal view for given data. The hexadecimal view is made up of one or more rows. Every row except the last one represents 16 characters. Each row consists of three columns separated by a space:

• addr: the 4-digit hexadecimal beginning address of this row.
• dump: the hexadecimal representation of this row, separating every two characters by a whitespace. If there are less than 16 characters in the last row, pad it with spaces.
• text: the ASCII translation of this row, with uppercase characters converted to lowercase and lowercase characters converted to uppercase.

Use lowercase for the letter digits. See sample for more details.

Input

There are multiple test cases. Each line is a test case. The line is made up of no less than 1 and no more than 4096 printable characters including spaces.

Output

For each test case, output its hexadecimal view. Do not output any extra spaces after the last character of text.

Sample Input

Hex Dump
#include <cstdio>
printf("Hello, World!\n");
main = do getLine >>= print . sum . map read . words

Sample Output

0000: 4865 7820 4475 6d70                     hEX dUMP
0000: 2369 6e63 6c75 6465 203c 6373 7464 696f #INCLUDE <CSTDIO
0010: 3e                                      >
0000: 7072 696e 7466 2822 4865 6c6c 6f2c 2057 PRINTF("hELLO, w
0010: 6f72 6c64 215c 6e22 293b                ORLD!\N");
0000: 6d61 696e 203d 2064 6f20 6765 744c 696e MAIN = DO GETlIN
0010: 6520 3e3e 3d20 7072 696e 7420 2e20 7375 E >>= PRINT . SU
0020: 6d20 2e20 6d61 7020 7265 6164 202e 2077 M . MAP READ . W
0030: 6f72 6473                               ORDS

Author: WU, Zejun

題目比較簡單，輸入輸出注意點。。。

看來要去復習下C++的輸入輸出的函數了。。。。。

WR了好多次，原來數組太小。題目給的是4096的，我卡的太緊了，，，下次開大點了。。。

C語言的函數輕松過掉

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<iomanip>
using namespace std;
char str[5000];
int main()
{
    int t=0;
    int len;
    while(cin.getline(str,4100))//這里要開大點，錯了好幾次了 
    {
        t=0;
        len=strlen(str);
        for(t=0;t<len;t+=16)
        {
             printf("%04x: ",t);
             for(int i=t;i<t+16;i+=2)
             {
                 if(i<len)  printf("%02x",str[i]);
                 else  printf("  ");
                 if(i+1<len) printf("%02x ",str[i+1]);
                 else  printf("   ");
             }    
             for(int i=t;i<t+16&&i<len;i++)
             {
                 if(str[i]>='a'&&str[i]<='z') printf("%c",str[i]-'a'+'A');
                 else if(str[i]>='A'&&str[i]<='Z')  printf("%c",str[i]-'A'+'a');
                 else printf("%c",str[i]);
             }    
            printf("\n");
         }    
    }  
    return 0;  
}    

Counting Game 

There are n people standing in a line, playing a famous game called ``counting". When the game begins, the leftmost person says ``1" loudly, then the second person (people are numbered 1 to n from left to right) says ``2" loudly. This is followed by the 3rd person saying ``3" and the 4th person say ``4", and so on. When the n-th person (i.e. the rightmost person) said ``n" loudly, the next turn goes to his immediate left person (i.e. the (n - 1)-th person), who should say ``n + 1" loudly, then the (n - 2)-th person should say ``n + 2" loudly. After the leftmost person spoke again, the counting goes right again.

There is a catch, though (otherwise, the game would be very boring!): if a person should say a number who is a multiple of 7, or its decimal representation contains the digit 7, he should clap instead! The following tables shows us the counting process for n = 4(`X' represents a clap). When the 3rd person claps for the 4th time, he's actually counting 35.

Given n, m and k, your task is to find out, when the m-th person claps for the k-th time, what is the actual number being counted.

Input 

There will be at most 10 test cases in the input. Each test case contains three integers n, m and k (

2n100,

1mn,

1k100) in a single line. The last test case is followed by a line with n = m = k = 0, which should not be processed.

Output 

For each line, print the actual number being counted, when the m-th person claps for the k-th time. If this can never happen, print `-1'.

Sample Input 

4 3 1
4 3 2
4 3 3
4 3 4
0 0 0

Sample Output 

17
21
27
35

The Seventh Hunan Collegiate Programming Contest
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

#include<stdio.h>
#include<string.h>
bool aa[1000000];
bool solve(int n)
{
    if(n%7==0)  return true;
    int t=n;
    while(t)
    {
        if(t%10==7)  return true;
        t/=10;
    }    
    return false;
}  
void calc()
{
    for(int i=1;i<1000000;i++)
    {
        if(solve(i))  aa[i]=true;
        else  aa[i]=false;
    }    
    
}      
int main()
{
    int n,m,k;
    int t;
    int cnt;
    calc();
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        if(n==0&&m==0&&k==0) break;
        t=m;
        cnt=0;
        while(1)
        {
            if(aa[t])cnt++;
            if(cnt==k)
            {
                printf("%d\n",t);
                break;
            }  
            if(n==m)
            {
                t+=2*(m-1); 
                continue;
            }  
            if(m==1)
            {
                t+=2*(n-m);
                continue;
            }      
            t+=2*(n-m);
            if(aa[t])cnt++;
            if(cnt==k)
            {
                printf("%d\n",t);
                break;
            }  
               
            t+=2*(m-1);  
        }    
    } 
    return 0;   
}    

One-Two-Three 

Your little brother has just learnt to write one, two and three, in English. He has written a lot of those words in a paper, your task is to recognize them. Note that your little brother is only a child, so he may make small mistakes: for each word, there might be at most one wrong letter. The word length is always correct. It is guaranteed that each letter he wrote is in lower-case, and each word he wrote has a unique interpretation.

Input 

The first line contains the number of words that your little brother has written. Each of the following lines contains a single word with all letters in lower-case. The words satisfy the constraints above: at most one letter might be wrong, but the word length is always correct. There will be at most 10 words in the input.

Output 

For each test case, print the numerical value of the word.

Sample Input 

3
owe
too
theee

Sample Output 

1
2
3

The Seventh Hunan Collegiate Programming Contest
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

#include<stdio.h>
#include<string.h>
int main()
{
    int n;
    char str[10];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",&str);
        int len=strlen(str);
        if(len>3) printf("3\n");
        else
        {
            if((str[0]=='o'&&str[1]=='n')||(str[0]=='o'&&str[2]=='e')||(str[1]=='n'&&str[2]=='e'))
               printf("1\n");
            else  printf("2\n");
        }    
    }  
    return 0;  
}    

Description

Input

Output

Sample Input

you are what you do

Sample Output

we are what we do

Source

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char str[1010];
int main()
{
    int i;
    int len;
    while(cin.getline(str,1010))
    {
        len=strlen(str);
        for(i=0;i<len;i++)
        {
            if(i+2<len&&str[i]=='y'&&str[i+1]=='o'&&str[i+2]=='u')
            {
                printf("we");
                i+=2;
                continue;
            }    
            printf("%c",str[i]);
        }    
        printf("\n");
    }    
    return 0;
}    

開學以來，各種忙碌，好久沒有刷題了。。。。今天在POJ刷了題，，，很簡單

Description

Input

Output

Sample Input

1/8+3/8
1/4-1/2
1/3-1/3

Sample Output

1/2
-1/4
0

Source

#include<stdio.h>
int gcd(int a,int b)
{
    if(b==0)  return a;
    return gcd(b,a%b);
}    
int lcm(int a,int b)
{
    int c=gcd(a,b);
    return a*b/c;
}   
int main()
{
    int a,b,c,d;
    char ch;
    while(scanf("%d/%d%c%d/%d",&a,&b,&ch,&c,&d)!=EOF)
    {
        int m=lcm(b,d);
        int n;
        if(ch=='+')   n=a*(m/b)+c*(m/d);
        else   n=a*(m/b)-c*(m/d);
        if(n==0) printf("0\n");
        else
        {
            int t=gcd(m,n);
            n=n/t;m=m/t;
            if(m<0)  m=-m,n=-n;
            if(m==1) printf("%d\n",n);
            else
            printf("%d/%d\n",n,m);
        }    
    }    
    return 0;
} 

Regular Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 682    Accepted Submission(s): 174

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
const double eps=1e-10;
double m[110][2];
int n;
const double PI=acos(-1.0);

int jug(double mid)
{
    double sum=0.0;
    int i;
    double temp;
    for(i=0;i<n;i++)
    {
        if(mid>(m[i][0]+m[i][1])-eps)  return 1;
        if(mid<fabs(m[i][0]-m[i][1])+eps)  return -1;
        temp=(m[i][0]*m[i][0]+m[i][1]*m[i][1]-mid*mid)/(2.0*m[i][0]*m[i][1]);
        sum+=acos(temp);
    }    
    if(sum>PI*2.0+eps)  return 1;
    if(sum<PI*2.0-eps) return -1;
    else  return 0;
}   
int main()
{
    int i,T;
    int iCase=0;
    scanf("%d",&T);
    double l,r,temp;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%lf",&m[i][1]);
            m[i+1][0]=m[i][1];
        }   
        m[0][0]=m[n][0];
        bool flag=false;
        double mid;
        int temp1;
        l=0;
        r=20000;
        while((r-l)>eps)
        {
            mid=(r+l)/2;
            temp1=jug(mid);
            if(temp1==0)
            {
                flag=true;
                break;
            }    
            if(temp1<0)  l=mid;
            else  r=mid;
        }  
        if(!flag) printf("Case %d: impossible\n",iCase);
        else  printf("Case %d: %.3lf\n",iCase,mid);  
    }  
    return 0;  
}    

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24579    Accepted Submission(s): 6774

#include<stdio.h>
#include<string.h>
const int MAXN=40000;//如果是10000的階乘，改為40000就夠了 
int f[MAXN];
int main()
{
    int i,j,n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(f,0,sizeof(f));
        f[0]=1;
        for(i=2;i<=n;i++)
        {
            int c=0;
            for(j=0;j<MAXN;j++)
            {
                int s=f[j]*i+c;
                f[j]=s%10;
                c=s/10;
            }    
        } 
        for(j=MAXN-1;j>=0;j--) 
           if(f[j]) break;//忽略前導0
        for(i=j;i>=0;i--)  printf("%d",f[i]);
        printf("\n");   
    }    
    return 0;
}    

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4004

本文作者 ：kuangbin   @SHU

博客：www.cnblogs.com/kuangbin

題意：一條長為l的河上有n個石頭，告訴每個石頭的位置，一個青蛙從頭跳，求最少每次跳多少距離才能跳到對面。最多跳m次

解題策略：二分答案。。。其實二分的效率是很高的噢~~~~很常用的算法。。

直接貼代碼了：

/*HDU 4004
二分 
*/
#include<stdio.h>
#include<algorithm>
using namespace std;
#define MAXN 500010
const int INF=0x7ffffff;
int d[MAXN];
int L;
int calc(int x)//這個函數是計算對于跳躍能力是x的青蛙，最少跳幾次過河
{
    int now=0;
    int cnt=0;
    int ps=0;
    while(now<L)
    {
        now+=x;
        while(now>=d[ps+1]) ps++;
        now=d[ps];
        cnt++;
    }    
    return cnt;
}
         
int main()
{
    int n,m;
    int left,right,mid;
    int minx;//兩個石頭間的最大距離，就是青蛙最小的跳的能力
    while(scanf("%d%d%d",&L,&n,&m)!=EOF)
    {
        minx=0;
        for(int i=1;i<=n;i++)
           scanf("%d",&d[i]);
        sort(d+1,d+n+1);
        d[0]=0;
        d[n+1]=L;
        d[n+2]=INF;
        for(int i=1;i<=n+1;i++)
           if(minx<d[i]-d[i-1])
             minx=d[i]-d[i-1];
        left=minx;right=L;
        int res;
        while(left<=right)
        {
            mid=(right+left)>>1;
            if(calc(mid)<=m)
            {
                res=mid;
                right=mid-1;
            }    
            else  left=mid+1;
        }    
        printf("%d\n",res);
    }     
    return 0;
}    

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

Output

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

Source

Southeastern Europe 2003

題目鏈接:http://poj.org/problem?id=1459

writed by kuangbin

轉載請注明出處

直接套用最大流的模板的，主要是建圖的過程。

輸入分別為m個點，a個發電站，b個用戶，n條邊；接下去是n條邊的信息(u,v)cost，cost表示邊(u,v)的最大流量；a個發電站的信息(u)cost，cost表示發電站u能提供的最大流量；b個用戶的信息(v)cost，cost表示每個用戶v能接受的最大流量。
    典型的最大網絡流中多源多匯的問題，在圖中添加1個源點S和匯點T，將S和每個發電站相連，邊的權值是發電站能提供的最大流量；將每個用戶和T相連，邊的權值是每個用戶能接受的最大流量。從而轉化成了一般的最大網絡流問題，然后求解。

 

具體看程序吧，不解釋.

/**//*
POJ 1459
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN=110;
const int INF=0x7fffffff;
int map[MAXN][MAXN],path[MAXN],flow[MAXN],start,end;
int n;//點的個數
queue<int>q;
int bfs()
{
    int i,t;
    while(!q.empty()) q.pop();//清空隊列
    memset(path,-1,sizeof(path));
    path[start]=0;
    flow[start]=INF;
    q.push(start);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t==end)  break;
        for(i=0;i<=n;i++)
        {
            if(i!=start&&path[i]==-1&&map[t][i])
            {
                flow[i]=flow[t]<map[t][i]?flow[t]:map[t][i];
                q.push(i);
                path[i]=t;
            }    
        }   
           
    }  
    if(path[end]==-1)  return -1;
        return flow[n];     
}       
int Edmonds_Karp()
{
    int max_flow=0,step,now,pre;
    while((step=bfs())!=-1)
    {
        max_flow+=step;
        now=end;
        while(now!=start)
        {
            pre=path[now];
            map[pre][now]-=step;
            map[now][pre]+=step;
            now=pre;
        }    
    }  
    return max_flow;  
}    
int main()
{
    int i,u,v,z,np,nc,m;
    while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
    {
        memset(map,0,sizeof(map));
        while(m--)
        {
            while(getchar()!='(');
            scanf("%d,%d)%d",&u,&v,&z);
            u++;v++;
            map[u][v]=z;
        }    
        while(np--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
            u++;
            map[0][u]=z;
        }  
        while(nc--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
            u++;
            map[u][n+1]=z;
        }  
        n++;
        start=0;end=n;
        printf("%d\n",Edmonds_Karp());    
    }    
    return 0;
}  

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