codeforces 552C Vanya and Scales 【思维】

题目链接：http://codeforces.com/contest/552/problem/C

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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.

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题目大意 ：

就是有w^0,w^1,W^2…..w^100 ,这么些个砝码 每种砝码各一个 问用这些砝码能不能称出m质量

解题思路：

就是正常使用天平 模拟下就好了

注意的是砝码可以放到天平的右边 且每种砝码只有一个

这样的话 只要判断是不是有那种砝码使用超过1次就行了

判断的时候其实就是把m写成w进制数就好了

每一数位上的数字就是对应的砝码用了多少次 所以只可能为 0 or 1 or w-1；

0和1 很好理解

为啥好友w-1呢

其实这么想 就是w^i 和w^(i+1)组成的 就是这两个放在的两边而已

所以只要这么判断就行了

for(int i=0; i<k; i++)
{
if(a[i]>=w)   a[i]-=w,a[i+1]++;//!!!注意这个  以为下面一行给高位+1  所以可能会==w  一定不能少判断这个 因为这个WA2发....
if(a[i]==w-1) a[i]=0,a[i+1]++;
if(a[i]>1)  flag = 0;
}

附本题代码

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#include <bits/stdc++.h>
using namespace std;
typedef long long int LL ;
#define INF 0x3f3f3f3f
const int M = 1e5+12;

int  a[101];

int main()
{
ios::sync_with_stdio(false);
int w,m;
while(cin>>w>>m)
{
if(w==2) {puts("YES");continue; }

bool flag = 1;
for(int i=0; i<101; i++)   a[i]=0;

int k=0;
while(m)
{
a[k++]=m%w;
m/=w;
}

for(int i=0; i<k; i++)
{
if(a[i]>=w)   a[i]-=w,a[i+1]++;
if(a[i]==w-1) a[i]=0,a[i+1]++;

if(a[i]>1)  flag = 0;
}

if(flag)    puts("YES");
else        puts("NO");
}
return 0;
}