HDU 5373 The shortest problem（同余模定理）

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 135    Accepted Submission(s): 74

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

Source

2015 Multi-University Training Contest 7

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<stack>
#include<queue>
#include<map>

#define eps 1e-6
#define INF 0x3f3f3f3f

using namespace std;

int a[1000010];
int n,m;
char str[10001001];
int k;

int main()
{
    k = 0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n == -1 && m == -1)
        {
            break;
        }
        int sum = 0;
        a[0] = n;
        for(int i=1;i<=m;i++)
        {
            int t = n;
            int ans = 0;
            while(t)
            {
                ans += t%10;
                t = t/10;
            }
            a[i] = sum + ans;
            sum += ans;
            n = sum;
        }
        int h = 0;
        for(int i=m;i>=0;i--)
        {
            int pp = a[i];
            while(pp>0)
            {
                str[h++]=pp%10+'0';
                pp = pp/10;
            }
        }
        str[h] = '\0';
        int cnt = 0;
        for(int i=h-1;i>=0;i--)
        {
            cnt = cnt * 10 + (str[i]-'0');
            cnt = cnt%11;
        }
        printf("Case #%d: ",++k);
        if(cnt == 0)
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}