Beans Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2845 Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime,…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3521    Accepted Submission(s): 1681 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2596    Accepted Submission(s): 1279 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2518    Accepted Submission(s): 1250 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled wi…

题意:容易理解. 分析:以后碰到这种类型的题,就要考虑把矩阵先按行来处理,再按列处理.先算出每行能够能够得到的最大值,然后按列处理即可. 代码实现: #include<stdio.h> #include<string.h> int n,m; ],dp[][],b[]; int max(int x,int y) { return x>y?x:y; } int main() { int i,j,res; while(scanf("%d%d",&n,&a…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyo…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4456    Accepted Submission(s): 2105 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled w…

Beans Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3418    Accepted Submission(s): 1629 Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled wi…

/*递推公式dp[i]=MAX(dp[i-1],dp[i-2]+a[j])*/ #include<stdio.h> #include<string.h> #define N 210000 int a[N],f[N],dp[N]; int Max(int v,int vv) { return v>vv?v:vv; } int main() { int n,m,i,j,k; while(scanf("%d%d",&n,&m)!=EOF) { m…

#include<stdio.h> #define N 200100 int f[N]; int  a[N],n; int main() { int m,j,i,suma,sumb,sumc,sumd; while(scanf("%d%d",&m,&n)!=EOF) {        for(i=1;i<=m;i++)   for(j=1;j<=n;j++)   scanf("%d",&f[(i-1)*n+j]);  …

DP 158:11:22 1205:00:00   Overview Problem Status Rank (56) Discuss Current Time: 2015-11-26 19:11:23 Contest Type: Private Start Time: 2015-11-20 05:00:00 Contest Status: Running End Time: 2016-01-09 10:00:00 Manager: qwerqqq Clone this contest Edit…

意甲冠军: 1. 对于每一行是,对不相邻的同一时间数取: 2.它是相同的列,相邻行不能同时服用: 3.因此,我们可以得到状态方程:dp[i]=dp[i-1]>(dp[i-2]+a[i])?dp[i-1]:dp[i-2]+a[i].先对每一行运用,在对每一行求出的和作为一组运用.可得终于结果. #include<iostream> using namespace std; int col[200001]; int dp[200001]; int GetMaxRow(int a[],int…

http://acm.hdu.edu.cn/showproblem.php?pid=2844 题意: 有n个硬币,知道其价值A1.....An.数量C1...Cn.问在1到m价值之间,最多能组成多少种价值. 思路: dp[i]表示i价值能够组成的最大种数. New~ 欢迎“热爱编程”的高考少年——报考杭州电子科技大学计算机学院关于2015年杭电ACM暑期集训队的选拔 Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327…

1.Robberies 连接 :http://acm.hdu.edu.cn/showproblem.php?pid=2955     背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);    正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q[i…

HDU 动态规划(46道题目)倾情奉献~ [只提供思路与状态转移方程] Robberies http://acm.hdu.edu.cn/showproblem.php?pid=2955      背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);          正确的…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2845 题意:吃豆子游戏 , 当你吃了一个格子的豆子 , 该格子左右两个和上下两行就不能吃了 , 输入每个格子的豆子数 , 求你最多能吃多少颗豆子? 在最大连续数列的基础上,改变了,求最大不连续和? 我们可以先求单独的每一行的的最大不连续和,相当于对矩阵进行压缩,将n列压缩成了一列.然后再求,这一列的最大不连续和. dp方程: m[j]=max(m[j-1],m[j-2]+temp) 对temp两种选择…

Repository Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 6444    Accepted Submission(s): 2096 Problem Description When you go shopping, you can search in repository for avalible merchandises b…

1.Robberies 连接 :http://acm.hdu.edu.cn/showproblem.php?pid=2955      背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和- 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);     正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q…

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…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

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:…

Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4569 Description Let f(x) = a nx n +...+ a 1x +a 0, in which a i (0 <= i <= n) are all known integers. We call f(x) 0 (mod…

The kth great number Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4006 Description Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a nu…

How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1796 Description   Now you get a number N, and a M-integers set, you should find out how many integers which are sm…

(转)http://blog.csdn.net/u013081425/article/details/39240021 http://acm.hdu.edu.cn/showproblem.php?pid=4418 读了一遍题后大体明白意思,但有些细节不太确定.就是当它处在i点处,它有1~m步可以走,但他走的方向不确定呢.后来想想这个方向是确定的,就是他走到i点的方向,它会继续朝着这个方向走,直到转向回头. 首先要解决的一个问题是处在i点处,它下一步该到哪个点.为了解决方向不确定的问题,将n个点转…

1.题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=3791 2.参考解题 http://blog.csdn.net/u013447865/article/details/22569639 这个题目本身简单,我的想法也很easy,但是发生在测试上,我把memset的参数搞错了,第三个是sizeof(a), 比如说int a[10],第三个参数应该是sizeof(10),也就是40,而我传的是10,导致后面的测试,都是答案错误,也就是后面的数据,初始…