hdu 5373 模拟

简单模拟题，可以利用一下能被11整除的数的特点：奇数位的数字和与偶数位的数字和之差能被11整除。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

const int N = 1000000;
int s[N];
int mid[N];

int main ()
{
    int n, t, _case = 0;
    while ( scanf("%d%d", &n, &t) != EOF )
    {
        if ( n == -1 && t == -1 ) break;
        int sum = 0, p = 0, cnt = 0;
        while ( n )
        {
            sum += n % 10;
            mid[p++] = n % 10;
            n /= 10;
        }
        for ( int i = p - 1; i >= 0; i-- )
        {
            s[cnt++] = mid[i];
        }
        while ( t-- )
        {
            int tmp = sum;
            p = 0;
            while ( tmp )
            {
                sum += tmp % 10;
                mid[p++] = tmp % 10;
                tmp /= 10;
            }
            for ( int i = p - 1; i >= 0; i-- )
            {
                s[cnt++] = mid[i];
            }
        }
        int ssum = 0;
        for ( int i = 0; i < cnt; i++ )
        {
            if ( i & 1 ) ssum += s[i];
            else ssum -= s[i];
        }
        _case++;
        if ( ssum % 11 == 0 )
        {
            printf("Case #%d: Yes
", _case);
        }
        else
        {
            printf("Case #%d: No
", _case);
        }
    }
    return 0;
}