题目描述
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.
Example
Input
1
10 3 10
1 2 3 4 5 6 7 8 9 10
Output
4
Note
Remember that k<=n<=16 and all numbers can be stored in 32-bit integer
题目思路：
对于这道题可以利用递归实现，为避免无效操作，我们每次从上一个加入count的数字后面开始遍历选择，即在递归时多传入一个参数记录当前数字对应的索引，下一次递归时从这个数字的下一位开始遍历递归。
当遍历到n层时，如果此时count刚好等于我们想要的那个值，ans++，否则，返回上一层。
代码实现：
#include<bits/stdc++.h>
using namespace std;
int n,k,s,ans;
int a[16];
bool flag=false;
void fun(int count,int kk,int d)//d代表上一层数对应的索引
{
if(kk==0)
{
if(count==s)
{
ans++;
return;
}
}
else{
for(int i=d;i<n;i++)
{
fun(count+a[i],kk-1,i+1);
}
}
}
int main()
{
int N;
scanf("%d",&N);
while(N--)
{
ans=0;
scanf("%d %d %d",&n,&k,&s);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
fun(0,k,0);
printf("%d\n",ans);
}
}
