C. Vanya and Scales （进制转换）

Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2(exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.

Input

The first line contains two integers w, m (2 ≤ w ≤ 109, 1 ≤ m ≤ 109) — the number defining the masses of the weights and the mass of the item.

Output

Print word 'YES' if the item can be weighted and 'NO' if it cannot.

    Sample test(s) 
  
      input 
    
3 7

      output 
    
YES

      input 
    
100 99

      output 
    
YES

      input 
    
100 50

      output 
    
NO

Note

Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

解题思路：把m转化为w进制，记M为m在w进制下的表示。

1、当M只由0、1组成时，明显，M 可由W进制表示，左边盘子放M ，右边盘子放相应1位置的砝码。

2、当M含有其它数字时，从低位到高位把非0、1的数字转化，使它的高位增加一，该位置数字减去W，

若值等于-1，则在M处再放置一个相应位的砝码；若值等于0（高位增一可为W），则不用放置。其它情况则为NO。

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define LL __int64
#define N 550
int a[N];
int main()
{
    LL w,m;
    int i,n;
    while(~scanf("%I64d%I64d",&w,&m))
    {
        n=0;
        while(m)
        {
            a[n++]=m%w;
            m/=w;
        }
        int flag=1;
        for(i=0;i<n;++i)
        {
            if(a[i]==0||a[i]==1)
                continue;
            if(i==n-1)
                ++n;
            ++a[i+1];
            a[i]-=w;
            if(a[i]==0||a[i]==-1)
                ;
            else
            {
                flag=0;
                break;
            }
        }
        printf("%s\n",(flag?"YES":"NO"));
    }
    return 0;
}

C. Vanya and Scales （进制转换）

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