Week3 A - 选数问题

Week3 A - 选数问题
Given n positive numbers, ZJM can select exactly K of them that sums to S. Now ZJM wonders how many ways to get it!

Input
The first line, an integer T<=100, indicates the number of test cases. For each case, there are two lines. The first line, three integers indicate n, K and S. The second line, n integers indicate the positive numbers.

Output
For each case, an integer indicate the answer in a independent line.
Note
Remember that k<=n<=16 and all numbers can be stored in 32-bit integer

Example
Input

1
10 3 10
1 2 3 4 5 6 7 8 9 10

Output

4

解题思路
DFS
递归+递归中判断是否达到目标 中途跳出(剪枝)

Code

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
using namespace std;
int st[16];
int n,k,s;
int ans;
void fff(int a,int sum,int t){//递归
	if(sum==s){//判断数值
		if(t==k){//判断相加次数
			ans++;
			return;
		}
		return;
	}
	for(int i=a;i<n;i++){
		sum+=st[i];
		fff(i+1,sum,t+1);
		sum-=st[i];
	}
}

int main(){
	int t;
	cin>>t;
	for(int i=0;i<t;i++){
		n=0;k=0;s=0;
		cin>>n>>k>>s;
		for(int y=0;y<n;y++){
			cin>>st[y];
		}
		//cout<<"cin finish"<<endl;
		for(int y=0;y<n-1;y++){//先排序 由小到大
			for(int u=0;u<n-y-1;u++){
				if(st[u]>st[u+1]){
					int temp=st[u];
					st[u]=st[u+1];
					st[u+1]=temp;
				}
			}
		}
		ans=0;
		fff(0,0,0);
		cout<<ans<<endl;
	}
	
}