（回溯法）数组中和为S的N个数

Given a list of numbers, find the number of tuples of size N that add to S.

for example in the list (10,5,-1,3,4,-6), the tuple of size 4 (-1,3,4,-6) adds to 0.

题目：

给一数组，求数组中和为S的N个数

思路：

回溯法，数组中每个数都有两种选择，取或者不取；

当选择的数等于N时，则判断该数之和是否等于S。

代码：

#include <iostream>
#include <vector>

using namespace std;

void GetSum(int sum,vector<int> &result,int *num,int n,int curSum,int index,int count){
if(count==n+1)
return;

if(index==4){
if(curSum==sum){
for(int i=0;i<4;i++)
cout<<result[i]<<" ";
cout<<endl;
}
return;
}

int x=num[count];
result.push_back(x);
GetSum(sum,result,num,n,curSum+x,index+1,count+1);
result.pop_back();
GetSum(sum,result,num,n,curSum,index,count+1);
}

int main()
{
int nums[]={10,-5,-5,0,5,4,6};
int len=sizeof(nums)/sizeof(nums[0]);
int curSum=0;
int index=0;
int count=0;
int sum=0;
vector<int> result;
GetSum(sum,result,nums,len,curSum,index,count);
return 0;
}