2015 Multi-University Training Contest 7 1005

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 380    Accepted Submission(s): 186

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

 

题意：将前面的每位数相加得到的数加到最后，如此往复执行n次，问最后得到的数字能否被11整除

分析：模拟一次就行，注意答案可能计算中的结果可能会超int

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
LL num,n,ans,sum;

LL calc(LL x)
{
    LL res=0;
    while(x)
    {
        res+=x%10;
        x=x/10;
    }
    return res;
}

LL count_digit(LL x)
{
    if(x==0) return 10;
    LL ok=1;
    while(x)
    {
        x=x/10;
        ok=ok*10;
    }
    return ok;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int Case=0;
    while(scanf("%I64d %I64d",&num,&n)!=EOF)
    {
        if(num==-1 && n==-1) break;

        ans=calc(num);//ans表示当前要加在后排的数
        sum=num;//sum表示累加的和
        for(int i=1;i<=n;i++)
        {
            sum=sum*count_digit(ans)+ans;
            sum=sum%11;
            ans=ans+calc(ans);
        }

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