HOJ1014

Niven Numbers

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	Source : Unknown
	Time limit : 1 sec		Memory limit : 32 M

Submitted : 5349, Accepted : 965

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.

Input

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

10 111
2 110
10 123
6 1000
8 2314
0

Sample Output

yes
yes
no
yes
no

本题意思为给定一个进制（base）,然后给一个在base进制下的数字NUM，判断NUM是否为尼玛数（NUM能否被NUM各位数字之和整除）

由于NUM的大小限制在int内，而base会让int型超出范围，比如8（10） = 1000（2），所以NUM需要为字符串类。第二个关键在于溢出处理，同HOJ1008中，
整除判断可以变求余边扩展。

#include<iostream>
using namespace std;
#include<string>

int main(){
    int base;    string num;
    while(cin>>base && base != 0){
        cin>>num;
        int x = 0;    int y = 0;
        for(int i = 0;i < num.length();i++){
            x += num[i] - '0';
        }
        for(int i = 0;i < num.length();i++){
            y = y * base + num[i] - '0';
            y %= x;
        }
        
        if(y == 0){
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
    return 0;
}