CF 552 C. Vanya and Scales

J - J

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Status

Practice

CodeForces 552C

Description

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 Input

Input

3 7

Output

YES

Input

100 99

Output

YES

Input

100 50

Output

NO

Hint

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.

一个砝码只能用一次。。。。

所以m%w==0||m%w==1||m%w==w-1就成立,,,

当m%w==w-1时，m=m / w 要加1；；

就如 100 9899 9899%100==99，，，，m=m/w=98,,,因为上面的是w-1，所以应该m++;

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int w,m;
int main()
{
while (~scanf("%d%d",&w,&m))
{
if (w==2) {
printf("YES\n");
continue;
}
bool fafe=true;
while (m)
{
if (m==1||m==w)
{
fafe=true;
break;
}
if (m%w!=1&&m%w!=0)
{   if (m%w!=w-1)
{
fafe=false;
break;
}
else
m+=w;
}
m/=w;
}
if (fafe)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}