hdu 5373 The shortest problem 2015多校联合训练赛#7 模拟

最新推荐文章于 2019-03-09 20:42:48 发布

原创最新推荐文章于 2019-03-09 20:42:48 发布 · 628 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#2015多校联合训练赛 #模拟

##ACM-ICPC编程题同时被 3 个专栏收录

173 篇文章

订阅专栏

ACM之2015多校联合训练赛

27 篇文章

订阅专栏

##模拟

5 篇文章

订阅专栏

本文介绍了一个有趣的数字游戏，通过累加数字并将其追加到原数末尾的方式，进行多次迭代，最终判断生成的数字是否能被11整除。通过模拟实现的方法，提供了一种简洁高效的解决方案。

The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 532 Accepted Submission(s): 271

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

Sample Output

  
   Case #1: Yes
Case #2: No

Source

2015 Multi-University Training Contest 7

对于s，算出s所有位上的数字之和f ，把f加入s后面，求t次操作后的数字能否被11整除。

模拟：s1记录得到数字对11的模数，s记录所有数字之和。

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

int main(){
    int n,t,tt=1;

    while(scanf("%d%d",&n,&t),n!=-1){
        int s = 0,s1=n%11;
        while(n){
            s += n%10;
            n /= 10;
        }
        for(int i = 0;i < t; i++){
            int f = s,f1=s;
            while(f){
                s += f%10;
                s1 = s1*10%11;
                f/=10;
            }
            s1 = (s1 + f1) % 11;
        }
        printf("Case #%d: ",tt++);
        if(s1 == 0) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}