fjnu 1894 Niven Numbers

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本文介绍了一种算法，用于判断给定的b进制数是否为Niven数。Niven数是指一个数能够被其各位数字之和整除的数。文章通过具体的代码示例详细解释了如何将输入的b进制数转换为十进制，并检查该数是否能被其各位数字之和整除。

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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

Sample Output

yes
yes
no
yes
no

KEY:这题不是什么难题……就是题目要理解清楚，给定一个b进制数num，转成10进制，然后除以它的位数和，看是否整除；

    那个输入让我贡献了3个wawhile（cin>>b>>num)是错的……

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;
    
}