https://codeforces.com/problemset/problem/552/C
思路:
Basically, let’s try to solve:
c0*w^0 + c1*w^1 + c2*w^2 + … = m
Where each c0, c1, c2 are either -1, 0 or 1.
-1 means we used the weight on right, +1 means we used the weight on left and 0 means we did not use the weight at all.
=> c1*w^1 + c2*w^2 + … = m - c0*w^0
=> w(c1 + c2*w^1 + c3*w^2 + … ) = m - c0
=> c1 + c2*w^1 + c3*w^2 + … = (m - c0)/w
For such a solution to exist (m — c0) must be a multiple of w.
还能/w取整就依次重复/w，然后重复过程。
Let’s divide both sides by w and recursively solve if we can find such an m-c0 where c0 = -1 or 0 or 1
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=2e5+1000;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
int main(void){
cin.tie(0);std::ios::sync_with_stdio(false);
LL w,m;cin>>w>>m;
bool flag=1;
while(m){
if((m-1)%w==0) m--;
else if((m+1)%w==0) m++;
else if(m%w!=0){
flag=0;
cout<<"NO"<<"\n";
break;
}
m/=w;
}
if(flag){
cout<<"YES"<<"\n";
}
return 0;
}