这个题的意思就是给出一个数m。以及一个以1为首元素。w为比例常数的等比数列,数列长度为101,数列中每一个数字最多仅仅能用一次。问是否存在xa+wb+……=wc+wd+……+we+m。

非常显然,换句话说就是问,是否存在m=wa+wb+……+wf-wc-wd-……-we。再进行化简就能够得到。是在问,是否存在m=((((wh±1)wi±1)±1)wj±1)wk……。

那么非常显然可以进行搜索,比方说用广搜,每次当前的m±1中若有因子w就所有约掉,然后丢到队列中,否则就不用管,仅仅要队列中可以出现1,那么这样的情况就是存在的。反之就是不存在。

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int cnt(int w,int x)
{
while(x%w==0)
x/=w;
return x;
}
queue<int>qq;
bool bfs(int w,int m)
{
int t=cnt(w,m);
if(t!=m)
qq.push(t);
else
qq.push(m);
while(qq.size())
{
t=qq.front();
qq.pop();
if(t==1)
return 1;
int t1=cnt(w,t+1);
if(t1!=t+1)
qq.push(t1);
t1=cnt(w,t-1);
if(t1!=t-1)
qq.push(t1);
}
return 0;
}
int main()
{
int w,m;
cin>>w>>m;
if(bfs(w,m))
puts("YES");
else
puts("NO");
}

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.

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