题目传送门:http://poj.org/problem?id=1191 棋盘分割 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16150   Accepted: 5768 Description 将一个8*8的棋盘进行如下分割:将原棋盘割下一块矩形棋盘并使剩下部分也是矩形,再将剩下的部分继续如此分割,这样割了(n-1)次后,连同最后剩下的矩形棋盘共有n块矩形棋盘.(每次切割都只能沿着棋盘格子的边进行) 原棋盘上每一格…

题目传送门:http://poj.org/problem?id=1579 Function Run Fun Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20560   Accepted: 10325 Description We all love recursion! Don't we? Consider a three-parameter recursive function w(a, b, c): if a <=…

Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for…

Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for…

Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for…

Description In Korea, the naughtiness of the cheonggaeguri, a small frog, is legendary. This is a well-deserved reputation, because the frogs jump through your rice paddy at night, flattening rice plants. In the morning, after noting which plants hav…

Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 11830   Accepted: 2814 Description Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It's hard to have a jo…

链接: https://vjudge.net/problem/POJ-3186 题意: FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.…

DP[i][j]表示现在开头是i物品,结尾是j物品的最大值,最后扫一遍dp[1][1]-dp[n][n]就可得到答案了 稍微想一下,就可以, #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm> #include<cstring> #include<cstring> #include<vect…

简单DP dp[i][j]表示的是i到j这段区间获得的a[i]*(j-i)+... ...+a[j-1]*(n-1)+a[j]*n最大值 那么[i,j]这个区间的最大值肯定是由[i+1,j]与[i,j-1]区间加上端点的较大值推过来的. #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<vector> #include<string>…