（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;
}