Vanya and Scales(思维)

最新推荐文章于 2019-06-09 21:50:15 发布

转载最新推荐文章于 2019-06-09 21:50:15 发布 · 136 阅读

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#数据结构与算法

探讨了如何通过加减特定权重使天平达到平衡的问题。解析了算法思路，并提供了实现代码，涉及进制转换与天平平衡条件判断。

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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.

题解：很棒的一个题，初一看完全没思路，到最后也没搞定。。。这个题的意思是给w的（0-100）次方，0-100次方只用一次，使天平平衡，

转化一下就是一个w进制的数能否通过加减w的（0-100）次方，0-100次方只用一次，使之等于0；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<algorithm>
using namespace std;
int a[110];
int main(){
    int w, m;
    while(~scanf("%d%d", &w, &m)){
        int k = 0; 
        while(m){
            a[k++] = m % w;
            m /= w;
        }
        int ans = 1;
        for(int i = 0; i < k; i++){
            if(a[i] == w){
                a[i] = 0;
                a[i + 1]++;
                continue;
            }
            if(a[i] <= 1){
                continue;
            }
            else if(a[i] == w - 1){
                a[i + 1]++;
                a[i] = 0;
                continue;
            }
            else
                ans = 0;
        }
        if(ans)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}