Codeforces Round #308 (Div. 2) C. Vanya and Scales dfs

C. Vanya and Scales

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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

3 7

Sample Output

YES

HINT

 

题意

给你100个砝码，第i个砝码质量是w^i，然后问你能不能在有m的情况下，左边和右边都放砝码，使得这个天平平衡

题解：

dfs直接暴力就好了……

对于这个砝码来说，只有3种选择，放左边，不放，放右边

由于2和3直接输出yes，所以答案也就没多少了

代码：

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-5
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************
ll w,m;
int flag;
void dfs(ll a,ll b,ll c)
{

    if(c==m)
    {
        flag=1;
        return;
    }
    if(flag||a==-1)
        return;
    if(abs(b)<=m*w)
    {
        dfs(1,b*w,c+b);
        dfs(1,b*w,c-b);
        dfs(1,b*w,c);
    }

    return;
}
int main()
{
    flag=0;
    w=read(),m=read();
    if(w<=3)
    {
        cout<<"YES"<<endl;
        return 0;
    }
    dfs(1,1,0);
    if(flag)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
}