周六周末组队训练赛. Dogs' Candies Time Limit: 30000/30000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3920    Accepted Submission(s): 941 Problem Description Far far away, there live a lot of dogs in the forest. Unlike…

Little Zu Chongzhi's Triangles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2515    Accepted Submission(s): 1427 Problem Description Zu Chongzhi (429–500) was a prominent Chinese mathematic…

Song Jiang's rank list Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2006    Accepted Submission(s): 1128 Problem Description <Shui Hu Zhuan>,also <Water Margin>was written by Shi Nai…

A Curious Matt Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3058    Accepted Submission(s): 1716 Problem Description There is a curious man called Matt. One day, Matt's best friend Ted is w…

Dogs' Candies Time Limit: 30000/30000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1701 Accepted Submission(s): 404 Problem Description Far far away, there live a lot of dogs in the forest. Unlike other dogs, thos…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5113 题目大意:有k种颜色的方块,每种颜色有ai个, 现在有n*m的矩阵, 问这k种颜色的方块能否使任意两个相连的方块颜色不一样填满整个矩阵,如果可以输出任意一种. 解题思路:由于n,m在[1, 5]内, 所以直接暴力枚举: 需要注意的是:需要剪枝一下. 代码如下: #include<stdio.h> #include<string.h> #include<cmath> #…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5115 题目大意:前面有n头狼并列排成一排, 每一头狼都有两个属性--基础攻击力和buff加成, 每一头狼的攻击力等于他自身的基础攻击力加上旁边狼对他的buff加成. 现在你被这群狼围着了, 必须将这些狼全部击败, 你才能逃出, 你每攻击一头狼,收到的伤害是狼的总攻击力.现在问,你逃出这些狼的围攻需要的最小代价(收到的伤害): 解题思路:声明dp数组, dp[i][j]代表从i到j的区间杀死所有狼的最…

A Simple Stone Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description After he has learned how to play Nim game, Bob begins to try another ston…

传送门 对于错想成lis的解法,提供一组反例 1 3 4 2 5同时对于这次案例也可以观察出解法:对于每一个数,如果存在比它小的数在它后面,它势必需要移动,因为只能小的数无法向右移动,而且每一次移动都必然可以使得这个数到达正确位置,这是根据题意而得的 #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<cstdlib> #include<io…

题目传送门 设dp[i][j]为杀掉区间i到j之间的狼需要付出的最小代价,那么dp[i][j]=min{dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]} Java代码 import java.io.*; import java.util.*; class MyInputStream extends InputStream { public BufferedInputStream bis = new BufferedInputStream(System.in);…