CodeForces - 552C Vanya and Scales

Posted 2020-12-01 cautx

K - Vanya and Scales

CodeForces - 552C

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.

Examples

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.

OJ-ID：
CodeForces-552C

author：
Caution_X

date of submission：
20191031

tags：
模拟

description modelling：
给定两个数m,w，现有一个天平，和若干个砝码，砝码质量w^0,w^1,w^2,w^3......w^n。
问能否通过质量m的物体和若干个砝码使得天平平衡

major steps to solve it：
（1）把数m转化成w进制，p[i]存w进制从右到左第i个位置的大小
（2）如果任意i满足p[i]<=1那么一定可以使得天平平衡（左边m，右边对应w进制为1的砝码w^i）
（3）对于p[i]>1的数就判断能否转化成p[i]<=1&&p[i+1]<=1,比如w进制第i位是w-1，而第i+1位是0
那么就可以写成p[i]=-1,p[i+1]=1,最后再判断是不是所有p[i]<=1即可（p[i]为-1说明w^i重的砝码和质量m物体放在了同一边）

warnings：
每个砝码只用一次

AC code:

#include<bits/stdc++.h>
using namespace std;
int p[105];
int main()
{
    int w,m;
    cin>>w>>m;
    int id=1;
    while(m) {
        p[id++]=m%w;
        m/=w;
    }
    if(id>100) {
        cout<<"NO"<<endl;
        return 0;
    }
    for(int i=1;i<id;i++) {
        if(p[i]>=w) {
            p[i+1]++;
            p[i]-=w;
        }
        if(p[i]<=1)    continue;
        else if(p[i]==w-1) {
            p[i]=0;
            p[i+1]++;
        }
        else {
            cout<<"NO"<<endl;
            return 0;
        }
    }
    cout<<"YES"<<endl;
    return 0;
}