AcWing 93. 递归实现组合型枚举

题目链接：点击这里

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<map>
#include<set>

using namespace std;
typedef long long ll;
const int MOD = 10000007;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 1010;
int n, m;

void dfs(int idx, int now, int state)	//state二进制数记录结果
{
	if(now + n - idx < m)	return;		//把后面的数都选上也达不到m个
	if(now==m)
	{
		for(int i = 0; i < n; ++i)
			if(state >> i & 1)			//state第i位是否为1 
				printf("%d ", i+1);		//因为是1~n，所以这里i+1 
		printf("\n");
		return;
	}
	//字典序较小的排在前面，所以，先选再不选
	dfs(idx + 1, now + 1, state|1<<idx);		//选idx，即state第idx位置成1
	dfs(idx + 1, now, state);					//不选idx
}

int main()
{
	scanf("%d%d", &n, &m);
	dfs(0, 0, 0);	
	return 0;
}