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 != ){
         cin>>num;
         int x = ;    int y = ;
         for(int i = ;i < num.length();i++){
             x += num[i] - '';
         }
         for(int i = ;i < num.length();i++){
             y = y * base + num[i] - '';
             y %= x;
         }

         if(y == ){
             printf("yes\n");
         }else{
             printf("no\n");
         }
     }
     return ;
 }