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  • HDU-5373 The shortest problem

    The shortest problem

    http://acm.hdu.edu.cn/showproblem.php?pid=5373

    Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 181 Accepted Submission(s): 92


    Problem Description
    In this problem, we should solve an interesting game. At first, we have an integer n, then we begin to make some funny change. We sum up every digit of the n, then insert it to the tail of the number n, then let the new number be the interesting number n. repeat it for t times. When n=123 and t=3 then we can get 123->1236->123612->12361215.
     
    Input
    Multiple input.
    We have two integer n (0<=n<=104 ) , t(0<=t<=105 ) in each row.
    When n==-1 and t==-1 mean the end of input.
     
    Output
    For each input , if the final number are divisible by 11, output “Yes”, else output ”No”. without quote.
     
    Sample Input
    35 2
    35 1
    -1 -1
     
    Sample Output
    Case #1: Yes
    Case #2: No
     
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <malloc.h>
    #define Max(a,b) (a>b?a:b)
    #define Min(a,b) (a<b?a:b)
    #define MAX 999999999
    #define LL long long
    #define M 6666666
    using namespace std;
    int main()
    {
        int n,m,i,j,k,l,cas=1;
        while(~scanf("%d%d",&n,&m))
        {
            if(n==-1&&m==-1)break;
             if(m==0)
             {
                  if(n%11==0)printf("Case #%d: Yes
    ",cas++);
                else printf("Case #%d: No
    ",cas++);
                continue;
             }
            k=n;
            l=0;
            while(k>0)
            {
                l+=k%10;
                k/=10;
            }//cout<<l<<endl;
            while(m--)
            {
                k=l;//cout<<l<<" ";
                while(k>0)
                {
                    k/=10;
                    n*=10;
                }
                n=(n+l)%11;
                k=l;
                while(k>0)
                {
                    l+=k%10;
                    k/=10;//cout<<l<<endl;
                }
            }
            if(n==0)printf("Case #%d: Yes
    ",cas++);
            else printf("Case #%d: No
    ",cas++);
        }
        return 0;
    }
    Source
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  • 原文地址:https://www.cnblogs.com/cancangood/p/4721864.html
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