fjnu 1894 Niven Numbers

Description

A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.

Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.

Output

For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.

Sample Input

110 1112 11010 1236 10008 23140

Sample Output

yesyesnoyesno 

KEY:先将Base进制的数转成10进制的，然后与他的原来的位数求余就行了；

Source:

#include<iostream.h>
#include<math.h>

int judge(int base,char num[])
...{
    long sum1=0,sum2=0;
    int len=strlen(num);
    int i;
    for(i=len-1;i>=0;i--)
    ...{
        sum1+=pow(base,len-i-1)*(num[i]-'0');
        sum2+=num[i]-'0';
    }
    if(sum1%sum2==0) return 1;
    else return 0;
}

int main()
...{
//    freopen("fjnu_1894.in","r",stdin);
    int b;
    int N;
    cin>>N;
    char num[100];
    for(int i=1;i<=N;i++)
    ...{
        while(cin>>b)
        ...{
            if(b==0) break;
            cin>>num;
            if(judge(b,num)) cout<<"yes"<<endl;
            else cout<<"no"<<endl;
        }
        if(i!=N) cout<<endl;
    }
     return 0;
    
}