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…

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[…

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…

题意:给你一些N个点,M条边,走每条边要花费金钱,然后给出其中必须访问的点,在这些点可以打工,但是需要先拿到证书,只可以打一次,也可以选择不打工之直接经过它.一个人从1号点出发,给出初始金钱,问你能不能访问所以的点,并且获得所以证书. 题解:目标是那些一定要访问的点,怎么到达的我们不关心,但是我们关心花费最少的路径,而且到达那个点后是一定要打工的,如果只是经过,那么在求花费最少的路径的时候已经考虑过了. 因此先用Folyd求出各个点直接的最短路径,由于N很小,又只要求出一个解,所以直接dfs暴搜…

Zero SumConsider the sequence of digits from 1 through N (where N=9) in increasing order: 1 2 3 ... N. Now insert either a `+' for addition or a `-' for subtraction or a ` ' [blank] to run the digits together between each pair of digits (not in front…