Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.I. M. Hei, the lead cow pasture architect, is in charge of creating a…

USACO ORZ Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2309 Accepted Submission(s): 826 Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. Th…

USACO ORZ Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geom…

USACO ORZ Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3581    Accepted Submission(s): 1196 Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastur…

原题代号:HDU 4277 原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4277 原题描述: USACO ORZ Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5208    Accepted Submission(s): 1725 Problem Description Like ev…

USACO ORZ Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3809    Accepted Submission(s): 1264 Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastu…

Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.I. M. Hei, the lead cow pasture architect, is in charge of creating a…

题目大意:有N个木棒,相互组合拼接,能组成多少种不同的三角形. 思路:假设c>=b>=a 然后枚举C,在C的dfs里嵌套枚举B的DFS. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #define mod 2000007 using namespace std; int n; int X[…

题意: 给你n个数,要你用光所有数字组成一个三角形,问能组成多少种不同的三角形 时间分析: 3^15左右 #include<stdio.h> #include<set> using namespace std; set<long long>s; int _case,n,sum; ]; int Dfs(int a,int b,int c,int m) { if(m==n) { if(a>b||b>c) ; if(a&&b&&c&…

没什么好方法,只能用dfs了. 代码如下: #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> #include<set> #define I(x) scanf("%d",&x) using namespace std; ],n,ans; set<pair<int,in…