当前位置：网站首页>Counting garlic guest: the solution of the equation

Counting garlic guest: the solution of the equation

2022-04-23 01:25:00 【OldLeft】

Insert picture description here
Insert picture description here
The sample input ：

3
100
1 2
-1 2
1 2

Sample output

104

Answer key ：
Adopt the idea of violence , Enumerate each x All possible values , Then judge whether the calculation result is 0. Pay attention to pretreatment , Otherwise it will time out .

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,ans;
int k[10],p[10];
ll x[10][150];
void dfs(int ind,ll s){
    
    if(ind>n) return;
    if(ind==n){
    // Recursive export 
        if(s==0){
    // Check whether the calculation result is 0
            ans++;
        }
        return;
    }
    for(int i=1;i<=m;i++){
    // Enumeration solution 
        dfs(ind+1,s+x[ind][i]);
    }
}
int main(){
    
    cin>>n;
    cin>>m;
    for(int i=0;i<n;i++){
    
        cin>>k[i]>>p[i];
        for(int j=1;j<=m;j++){
    
            x[i][j]=k[i]*pow(j,p[i]);// Preprocessing 
        }
    }
    dfs(0,0);
    cout<<ans<<endl;
    return 0;
}

版权声明
本文为[OldLeft]所创，转载请带上原文链接，感谢
https://yzsam.com/2022/04/202204230120462576.html

边栏推荐

猜你喜欢

随机推荐