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);
}