CF 552-C. Vanya and Scales

http://codeforces.com/problemset/problem/552/C

C. Vanya and Scales

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 than2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with massm 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 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 masses9 and
1, correspondingly. Then7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass99 and the weight of mass
1, and the second pan can have the weight of mass100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

一个天平，100个weights，重量为w的0次幂到100次幂 各一个， 通过天平和一些weights，问测出重量为m的物体是否可行。

方法： 将m转化成w进制的数。 由于每种砝码只有1个。所以各个位如果 是0，代表该砝码没用到， 1代表砝码与物体异侧，w-1代表砝码与物体同侧。其他情况则不能构成m

#include<bits/stdc++.h>
#define For(i,a,b) for(int (i)=(a);(i) < (b);(i)++)
#define rof(i,a,b) for(int (i)=(a);(i) > (b);(i)--)
#define IOS ios::sync_with_stdio(false)
#define lson l,m,rt <<1
#define rson m+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

const int maxn = 2e2+10;
const int INF =0x3f3f3f3f;
ll w,m;
int main()
{
cin>>w>>m;
ll bit[maxn];
mem(bit,0);
int len=0;
while(m){
bit[len++]=m%w;
m/=w;
}
For(i,0,len){
if(bit[i]==0||bit[i]==1) continue;
if(bit[i]>=w-1) {
bit[i+1]++;len++;
continue;
}
else{
cout<<"NO"<<endl;
return 0;
}
}
cout<<"YES"<<endl;
return 0;
}