/* 动态转移方程:dp[i][j]=max(dp[i-1]+a[i], max(dp[t][j-1])+a[i]) (j-1<=t<i) 表示的是前i个数j个字段和的最大值是多少! */ 1 #include<iostream> #include<cstdio> #include<cstring> #define N 10000 using namespace std; int dp[N][N], num[N]; int main() { int n, m…

HDU 1176 免费馅饼 (动态规划) Description 都说天上不会掉馅饼,但有一天gameboy正走在回家的小径上,忽然天上掉下大把大把的馅饼.说来gameboy的人品实在是太好了,这馅饼别处都不掉,就掉落在他身旁的10米范围内.馅饼如果掉在了地上当然就不能吃了,所以gameboy马上卸下身上的背包去接.但由于小径两侧都不能站人,所以他只能在小径上接.由于gameboy平时老呆在房间里玩游戏,虽然在游戏中是个身手敏捷的高手,但在现实中运动神经特别迟钝,每秒种只有在移动不超过一米的范围…

HDU 1074 Doing Homework (动态规划,位运算) Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the d…

HDOJ(HDU).1258 Sum It Up (DFS) [从零开始DFS(6)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DF…

Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequ…

题目大意: 给你一个总和(total)和一列(list)整数,共n个整数,要求用这些整数相加,使相加的结果等于total,找出所有不相同的拼凑方法. 例如,total = 4,n = 6,list = [4,3,2,2,1,1]. 有四种不同的方法使得它们相加的结果等于total(即等于4),分别为:4,3+1,2+2, 2+1+1. 在同一种拼凑方式中,每个数字不能被重复使用,但是在list中可能存在许多相等的数字. 输入: 输入包含许多测试用例,每个用例仅占一行.每个用例包含t(total)…

Sum Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=4704 Mean: 给定一个大整数N,求1到N中每个数的因式分解个数的总和. analyse: N可达10^100000,只能用数学方法来做. 首先想到的是找规律.通过枚举小数据来找规律,发现其实answer=pow(2,n-1); 分析到这问题就简单了.由于n非常大,所以这里要用到费马小定理:a^n ≡ a^(n%(m-1)) * a^(m-1)≡ a^(n%(m-…

C - 最大连续子序列 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1231 Appoint description: Description 给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj },其中 1 <= i <= j <= K.最大连续子…

Sum Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4704 Description   Sample Input 2   Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases. 题意…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15898    Accepted Submission(s): 5171 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

sum 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5776 Description Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO Input The first line of the input has an i…

Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5586 Description There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5586 Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 677    Accepted Submission(s): 358 Problem Description There is a number sequence A1,A2...…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a mor…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4432 代码: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int sum; int n,k; int tranfer(int num) { ; ) { int a = num%k; num = num/k; ret +…

http://acm.hdu.edu.cn/showproblem.php?pid=4407 题意:给定初始n个数1..n,两个操作,①1 x y p  询问第x个数到第y个数中与p互质的数的和; ②:2 x y  把第x个数变成y: 思路: 把p分解质因子,然后找出(1,pos)内与p不互质的,然后用的减去就是互质的和,第二个操作用到map映射,记录在那个位置改变之后的数. #include <cstdio> #include <cstring> #include <map…

Problem Description There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod10007.After that,the sum of n numbers should be as much as possible.What is the maximum s…

题目链接 Problem Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in th…

Sum Zero Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Problem Description There are 5 Integer Arrays and each of them contains no more than 300 integers whose value are between -100,000,000 and 100,000,000, You…

题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=4704 Problem Description   Sample Input 2 Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.   题意是输入一个N,求N被分成1个数的结果+被分成2个数的结果+...+被分成N个数的结果,N很大   1.隔板原…

题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=5776 Problem Description Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO Input The first line of the input has a…

Sum Of Gcd 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4676 Description Given you a sequence of number a1, a2, ..., an, which is a permutation of 1...n. You need to answer some queries, each with the following format: Give you two numbers L, R, y…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5433 Xiao Ming climbing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1346    Accepted Submission(s): 384 Problem Description Due to the curse m…

搬寝室 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14107    Accepted Submission(s): 4751 Problem Description 搬寝室是很累的,xhd深有体会.时间追述2006年7月9号,那天xhd迫于无奈要从27号楼搬到3号楼,因为10号要封楼了.看着寝室里的n件物品,xhd开始发呆,因为n是…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=4676 Sum Of Gcd Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 908    Accepted Submission(s): 438 Problem Description Given you a sequence of numb…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题目: Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a m…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1258 Sum It Up Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7758    Accepted Submission(s): 4067 Problem Description Given a specified total t a…

Max Sum Plus Plus 题意:题意理解了老半天,这里是说在给定数列中,取m组子数列,不能有重复,使得这些子序列的和最大: 就比如m=2时候,1 /2/-4/5/6.可以不用拿-4的意思: 思路:这道题的思路是动态规划,递推: 状态dp[i][j] 表示有前j个数,组成i组的和的最大值. 决策: 第j个数,要么包含在第i组里面,要么自己独立成组. 其中最后一组包含a[j].(这很关键) 则状态转移方程为:(在二维图中,就是要么从左边取,要么取上一行的最大值,下式中,左边max是包含在第…