B. Plus and Multiply

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已于 2022-10-28 10:34:09 修改

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分类专栏： codeforce 文章标签： c++

于 2022-10-27 16:54:20 首次发布

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博客围绕一个无限集合展开，该集合生成规则为：1在集合中，若x在集合中，x*a和x + b也在集合中。给定正整数a、b、n，需判断n是否在集合中。输入包含多组测试用例，输出根据判断结果为Yes或No，并给出了示例及推理思路。

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Problem - B - Codeforces

There is an infinite set generated as follows:

11 is in this set.
If xx is in this set, x⋅ax⋅a and x+bx+b both are in this set.

For example, when a=3a=3 and b=6b=6, the five smallest elements of the set are:

11,
33 (11 is in this set, so 1⋅a=31⋅a=3 is in this set),
77 (11 is in this set, so 1+b=71+b=7 is in this set),
99 (33 is in this set, so 3⋅a=93⋅a=9 is in this set),
1313 (77 is in this set, so 7+b=137+b=13 is in this set).

Given positive integers aa, bb, nn, determine if nn is in this set.

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤1051≤t≤105) — the number of test cases. The description of the test cases follows.

The only line describing each test case contains three integers nn, aa, bb (1≤n,a,b≤1091≤n,a,b≤109) separated by a single space.

Output

For each test case, print "Yes" if nn is in this set, and "No" otherwise. You can print each letter in any case.

Example

input

Copy

5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264

output

Copy

Yes
No
Yes
No
Yes

Note

In the first test case, 2424 is generated as follows:

11 is in this set, so 33 and 66 are in this set;
33 is in this set, so 99 and 88 are in this set;
88 is in this set, so 2424 and 1313 are in this set.

Thus we can see 2424 is in this set.

The five smallest elements of the set in the second test case is described in statements. We can see that 1010 isn't among them.

想出来的公式： $a^x+b*y=u$

x和y代表个数

$y=(u-a^x)/b$ 只要判断y是不是整数就行，也就是等式右边可以被整除。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<deque>
#include<cmath>
#include<string.h>
using namespace std;
// ctrl+shift+C 注释
//ctrl+shift+x 取消
#define int long long
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
const int N=2e5+10;
const ll M=1e18+10;
const int mod=1e9+7;
int aa[N],sum[N];
priority_queue<int,vector<int>,greater<int> >pq;
set<int>se;
map<int,int>mp;
queue<int>qu;
vector<int>v;
deque<int>de;
int u,a,b;
void solve()
{
    cin>>u>>a>>b;
    if(a==1)// 特判一下
    {
        if((u-1)%b==0)
        {
            cout<<"Yes"<<endl;
        }
        else cout<<"No"<<endl;
        return;
    }
    int r=log10(u)/log10(a);
   for(int i=0;pow(a,i)<=u;i++)
   {
      int res=u-pow(a,i);
       if(res%b==0)
       {
           cout<<"Yes"<<endl;
           return;
       }
   }
    cout<<"No"<<endl;
}
signed main()
{
   int t=1;
   cin>>t;
   while(t--)
   {
       solve();
   }
}

激动我居然写出来了qaq