CodeForces 552C Vanya and Scales

A - Vanya and Scales

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 552C

Description

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 massm 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 Input

Input

3 7

Output

YES

Input

100 99

Output

YES

Input

100 50

Output

NO

Hint

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.

---

FAQ | About Virtual Judge | Forum | Discuss | Open Source Projec

//题意：输入w，m。

表示给你一个w，你可以使用w^0,w^1,w^2w^3.........w^100;这一百零一个数，现在又给你一个数m，让你将m放到天平的右边，现在问你能否通过在两边天平上添加那101个砝码使得天平平衡。

求M=W(a)+W(b)+...+W(c)-W(d)-W(e)-....-W(f)其中W（a）为w的a次方且括号内的数不相同。

解题思路：将等式两边同时除以W若M%W！=0既考虑存在W（0）即考虑（M+1）%W和（M-1）%W若均不存在则无解否侧循环除以W当最终W为0时即有解否则无解

代码：

#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<cstring>#define max(a,b)(a>b?a:b)#define min(a,b)(a<b?a:b)#define INF 999999999using namespace std;typedef long long ll;int main(){    ll w,m;    while(scanf("%lld%lld",&w,&m)!=EOF)    {        while((m+1)%w==0 || (m-1)%w==0 || m%w==0)        {            if((m-1)%w==0)                m=(m-1)/w;            else if((m+1)%w==0)                m=(m+1)/w;            else if(m%w==0)                m=m/w;            if(m==0)                break;        }        if(m==0)            printf("YES\n");        else            printf("NO\n");    }    return 0;}