Codeforces 552C

C. Vanya and Scales
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

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.

题意
给出一个天平,天平一边有一个重为m的砝码.
给你若干的砝码,质量为w^0, w^1,….w^100(每种质量的砝码只有一个,问是否能使得天平平衡).

解题思路:
将m换成w进制的,从低位向高位遍历,如果是w-1,就添1并向前进位,如果是0或者1也可以.

因为每种质量的砝码都可以看做是w进制中的某一位.
那么如果当前位是0,则两边都不放,如果是1,则可以在m的另一边放上一个等效的砝码,如果是w-1,那么可以是m边为质量是w^(w-2),另一边是w^(w-1)的等效构成.

AC代码:

#include <bits/stdc++.h>using namespace std;vector<int> a;int main(){    ios::sync_with_stdio(0);    int w, m;    cin >> w >> m;    while(m)        a.push_back(m % w), m /= w;    a.push_back(0);//    for(int i = 0; i < a.size(); i++)//        cout << a[i] << endl;    for(int i = 0; i < a.size(); i++)    {        if(a[i] == 0 || a[i] == 1)            continue;        if(a[i] == w-1)        {            a[i] = 0;            a[i+1]++;            continue;        }        if(a[i] == w)        {            a[i] = 0;            a[i+1]++;            continue;        }        cout << "NO" << endl;        return 0;    }    cout << "YES" << endl;    return 0;}