The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 346 Accepted Submission(s): 167

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 1005

解题：由于能被11整除的数的特点是奇数位与偶数位的和的绝对值可以被11整除，所以，搞一下就可以了

 #include <bits/stdc++.h>

 using namespace std;

 int cur = ,n,m,d[];

 int solve(int x) {

     int a[] = {};

     cur = ;

     while(x) {

         a[cur^] += x%;

         x /= ;

         cur ^= ;

     }

     if(cur&) swap(d[],d[]);

     d[] += a[];

     d[] += a[];

     return d[] + d[];

 }

 int main() {

     int cs = ;

     while(scanf("%d%d",&n,&m),(~n) && (~m)) {

         d[] = d[] = ;

         for(int i = ; i <= m; ++i) n = solve(n);

         printf("Case #%d: %s\n",cs++,abs(d[]-d[])%?"No":"Yes");

     }

     return ;

 }

巴特西

2015 Multi-University Training Contest 7 hdu 5373 The shortest problem

The shortest problem

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