传送门 Description Dynamic Programming, short for DP, is the favorite of iSea. It is a method for solving complex problems by breaking them down into simpler sub-problems. It is applicable to problems exhibiting the properties of overlapping sub-problem…

Dynamic Programming? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Dynamic Programming, short for DP, is the favorite of iSea. It is a method for solving complex problems by breaking them down…

暴力: Problem : ( Dynamic Programming? ) Judge Status : Accepted RunId : Language : C++ Author : yudunfengqing Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta #include<stdio.h> #include<math.h> #include<algorithm> u…

//连续的和的绝对值最小 # include <stdio.h> # include <string.h> # include <algorithm> # include <math.h> using namespace std; int main() { int t,i,j,num,n,min1; int sum[1010],a[1010]; int cas=0; while(~scanf("%d",&t)) { while(t…

---恢复内容开始--- 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2962 Trucking Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11428    Accepted Submission(s): 1104 Problem Description A certain lo…

Trucking Problem Description A certain local trucking company would like to transport some goods on a cargo truck from one place to another. It is desirable to transport as much goods as possible each trip. Unfortunately, one cannot always use the ro…

King's Game 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5643 Description In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers, counterclockwise in a circle, in label 1,2,3...n. The firs…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…

http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格线满足两侧分别是海洋和陆地 这道题很神 首先考虑一下,什么情况下能够对答案做出贡献 就是相邻的两块不一样的时候 这样我们可以建立最小割模型,可是都说是最小割了 无法求出最大的不相同的东西 所以我们考虑转化,用总的配对数目 - 最小的相同的对数 至于最小的相同的对数怎么算呢? 我们考虑这样的构造方法:…