同余模定理 HDOJ 5373 The shortest problem

题目传送门

 /*
     题意：题目讲的很清楚：When n=123 and t=3 then we can get 123->1236->123612->12361215.要求t次操作后，能否被11整除
     同余模定理：每次操作将后缀值加到上次操作的值%11后的后面，有点绕，纸上模拟一下就行了
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-12 8:50:23
 * File Name     :E.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN];

 int part(int x) {
     int ret = ;
     while (x)   {
         ret += x % ;  x /= ;
     }
     return ret;
 }

 int cal(int x)  {
     int ret = ;
     while (x)   {
         ret *= ;  x /= ;
     }
     return ret;
 }

 int main(void)    {        //HDOJ 5373 The shortest problem
     int n, t, cas = ;
     while (scanf ("%d%d", &n, &t) == ) {
         if (n == - && t == -) break;

         a[] = n;   int sum = part (n);
         for (int i=; i<=t+; ++i)    {
             a[i] = a[i-] %  * cal (sum) + sum;
             sum += part (sum);
         }

         printf ("Case #%d: %s\n", ++cas, (a[t+] %  == ) ? "Yes" : "No");
     }

     return ;
 }