USACO ORZ Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2291    Accepted Submission(s): 822 Problem Description Like everyone, cows enjoy variety. Their current fancy is new shapes for pastur…

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…

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) 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…

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…

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&…

http://acm.hdu.edu.cn/showproblem.php?pid=5012 发现一个问题 假设Sting s = '1'+'2'+'3'; s!="123"!!!!!!  而是有乱码 先贴一份自己的TLE 代码, 超时应该是由于: 1.cin 2.map判重 map find太花时间 3.string花时间 4.事实上不用两个都旋转,仅仅要旋转一个即可,这样能够省下非常多时间 包含少用了make_pair pair的判重等等....哎  二笔了  太暴力了 #incl…