HDU 5313

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K 

(Java/Others)

Total Submission(s): 1484    Accepted Submission(s): 686

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 

//不要用sprintf  功能太多时间复杂度太大

#include <stdio.h>
#include <cmath>
using namespace std;

char s[1000011];
int k,an=0;
int sprint(char s[],int n)
{
    char ss[10000];
    int cnt=0;
    int t=0;
    while(n)
    {
        ss[cnt++]=n%10+'0';
        t+=n%10;
        n/=10;
    }
    for(int i=cnt-1;i>=0;i--)
        s[k++]=ss[i];
    s[k]='\0';
    return t;

}
int main()
{
    int n,m,cas=1;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1)
            break;
        k=0;
        int val=sprint(s,n);
        while(m--)
        {
            val+=sprint(s,val);
            //puts(s);
        }
        int sum1,sum2;
        sum1=sum2=0;
        for(int i=0;s[i]!='\0';i++)
        {
            if(i%2==1)
                sum1+=s[i]-'0';
            else
                sum2+=s[i]-'0';
        }
        if(int(abs(sum1-sum2))%11==0)
            printf("Case #%d: Yes\n",cas++);
        else
            printf("Case #%d: No\n",cas++);
    }
    return 0;
}