从n个数中选取m个数的所有组合

非递归实现：

#define N 7
#define M 3
int main()
{	int array[N] = { 1,2,3,4,5,6 ,7 };
	int i, j, k;

	for (i = 0; i<=N-M; i++)
		for (j = i + 1; j<=N-M+1; j++)
			for (k = j + 1; k<=N-M+2; k++)
				printf("(%d,%d,%d)\n", array[i], array[j], array[k]);
	return 0;
}

递归实现：从n个数中选取k个数，和为x

int n, k, x, num[100] = {2,3,3,4};
vector<int> temp;
void dfs(int index, int nowK, int sum) {
	if (nowK == k&&sum == x) {
		for (int i = 0; i < k; ++i) {
			printf("%d ", temp[i]);			
		}
		putchar('\n');
		return;
	}
	if (index >= n || nowK > k || sum > x)	return;
	temp.push_back(num[index]);
	dfs(index + 1, nowK + 1, sum + num[index]);
	temp.pop_back();

	dfs(index + 1, nowK, sum);
}