POJ 1664

这题真实水的无与伦比的快乐，当然中间debug稍微浪费了点时间。很简单的DFS

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <map>
#include <set>
using namespace std;

int n;
int DFS(int m, int cnt, int mx)
{
	if (n== cnt){
		if (m> mx){
			return 0;
		}
		return 1;
	}
	int ans= 0;
	for (int i= min(mx, m), j= m-i; i>= 0; --i, ++j){
		ans+= DFS(j, cnt+1, i);
	}

	return ans;
}

int main(int argc, char const *argv[])
{
	int kase;
	scanf("%d", &kase);

	while (kase--){
		int m;
		scanf("%d %d", &m, &n);
		printf("%d\n", DFS(m, 1, m));
	}

	return 0;
}

标签：cnt,1664,return,int,DFS,POJ,ans,include
来源： https://www.cnblogs.com/Idi0t-N3/p/14678998.html