The shortest problem（hdu，多校

最新推荐文章于 2025-08-11 09:20:14 发布

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最新推荐文章于 2025-08-11 09:20:14 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Tinamei/p/4721335.html

本博客探讨了一个有趣的数学游戏，通过不断求和并插入数字，最终判断结果是否能被11整除。详细介绍了游戏规则、输入输出方式，并提供了解决问题的代码示例。

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The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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<=

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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 8 //int main(int argc, char** argv) 
 9 
10 #define N 100005
11 
12 int main()
13 {
14     int n, m, d, a, sum, r, nu, t = 1;
15     
16     while(scanf("%d%d", &n, &m), n != -1 || m != -1)
17     {
18         r = n % 11;
19         sum = 0;
20         
21         while(n)   // 不知道为什么非得这么写才对~
22         {
23             sum += n % 10;
24             n /= 10;
25         }
26         
27         while(m--)
28         {
29             a = 1;
30             nu = n = sum;  // nu，存当前次数之前出现的根的总和
31             sum = 0;
32             
33             while(n)
34             {
35                 sum += n % 10;  // sum存目前的根总和的根
36                 n /= 10;
37                 a *= 10;
38             }
39             
40             r = (r * a + nu) % 11;
41             sum += nu;
42         }
43         
44         if(r == 0)
45             printf("Case #%d: Yes\n", t++);
46         else
47             printf("Case #%d: No\n", t++);
48     }
49     return 0;
50 }