2015 多校联赛 ——HDU5373（模拟）_in this problem, we should solve an interesting ga-优快云博客

本文链接：https://blog.youkuaiyun.com/Fun_Zero/article/details/47439973

本文介绍了一个基于整数变换的趣味数学游戏，并提供了一段C++实现代码。该算法通过不断累加数字并将其插入原数末尾的方式，生成新的数字，最终判断生成的数字是否能被11整除。

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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.

Sample Input

35 2
35 1
-1 -1

Sample Output

Case #1: Yes
Case #2: No

对生成的数%11，余数与下次多增的部分结合，不停求余

（官方：用long long 写不好的会超时，是int的四倍 - -）

#include <iostream>
#include <cstdio>

using namespace std;
typedef long long ll;

int f(int sum)
{
    int i = 0;
    while(sum>0)
    {
        sum /= 10;
        i++;
    }

    return i;
}

int p(int num)
{
    int i = f(num);
    int sum = 1;
    while(i--)
    {
        sum *= 10;
    }

    return sum;
}

int main()
{
    int n, t, tt=1;
    while(scanf("%d%d", &n, &t))
    {
        if(n==-1&&t==-1) break;

        int sum = 0;
        int div = 0;
        int temp;
        int temps;

        temp  = n;
        while(temp>0)
        {
            sum += (temp%10);
            temp /= 10;
        }

        for(int i=0; i<t; i++)
        {
            div = n % 11;

            n = div*p(sum) + sum;

            temp = sum;
            temps = 0;
            while(temp>0)
            {
                temps += (temp%10);
                temp /= 10;
            }
            sum += temps;

        }
        if(n%11==0) printf("Case #%d: Yes\n", tt++);
        else printf("Case #%d: No\n", tt++);
    }

    return 0;

}