CF 552C 进制转换

http://codeforces.com/problemset/problem/552/C

C. Vanya and Scales

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

/**
CF 552C 进制转换
题目大意：给定一个天平。砝码的重量为w的0~100次幂。每种砝码仅仅有一个，砝码能够放在左盘或者右盘。

给定物品的重量m。问是否有一种方案让天平两端平衡
解题思路：把m化为w进制。改进制数仅仅能有0,1或者w-1。若为w-1那么相当于在物品所放的盘里加一个砝码，然后在还有一盘加上w倍的砝码就可以。

直接w进制数当前位
          清0，将下一位+1就可以。最后看w进制数是否正好是一个01串就可以
*/
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
using namespace std;
int bit[40],n,m;
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int k=0;
        memset(bit,0,sizeof(bit));
        while(m)
        {
            bit[k++]=m%n;
            m/=n;
        }
        int flag=1;
        for(int i=0;i<k;i++)
        {
            if(bit[i]>=n)
            {
                bit[i]-=n;
                bit[i+1]++;
            }
            if(bit[i]==n-1)
            {
                bit[i]=0;
                bit[i+1]++;
            }
            else if(bit[i]>1)
            {
                flag=0;
                break;
            }
        }
        if(flag==0)
            puts("NO");
        else
            puts("YES");
    }
    return 0;
}