这是一个递归解决方案，用于查找集合的子集数（元素总和等于给定总和）。 为什么它只返回 1？

Question

//this approach is based on finding all the subsets of a set having n elements by finding subsets of n-1 elements. and so on through recursion..
int sum_subset(int set[], vector<int>& subset, int sum, int n, int l=0)
{
    //for  edge case of sum=0
    if (sum = 0) return 1;
    int s = 0;

    //base condition of recursion
    if (l == n) {
        for (int x:subset) {
            s += x;
        }
        return (s==sum)? 1: 0;
    }
    //calling the recursion for ->not including the nth element of the set of n elements in the subset of n-1 elements
    int a = sum_subset(set, subset, sum, n, l+1);

    //including nth element in the subset
    subset.push_back(set[l]);
    
    //calling the recursion for -> including the nth element of the set of n elements in the subset of n-1 elements
    int b = sum_subset(set, subset, sum, n, l+1);
    return a + b;
}  

int main() {
    int n;
    cout << "enter the size of set" << endl;
    cin >> n;
    int set[n];
    cout << "enter the set elements" << endl;

    for(int i = 0; i < n; i++) {
        cin >> set[i];
    }
    cout << "enter the sum " << endl;
    int sum;
    cin >> sum;

    vector<int> subset{0};
    int ans = sum_subset(set, subset, sum, n);
    cout << "subsets having sum =" << sum << "are->" << ans << endl;

}

//output
enter the size of set
5
enter the set elements
1
2
2
1
3
enter the sum 
3
subsets having sum =3are->1

Answer 1

您正在为sum变量分配值 0 而不是比较值：

    //for  edge case of sum=0
    if (sum = 0) return 1;
    int s = 0;

这本身隐藏了另一个问题：您永远不会从subset向量中删除值。 例如，第一次将值压入子集时，子集等于{3} 。 然后你只需添加越来越多的值，这个“子集”永远不会给你等于 3 的总和。

另一个不影响您的程序但很重要的问题。 在下面的行中，您没有创建空向量，而是创建了一个包含一个零元素的向量，该向量等于{0} ：

vector<int> subset{0};