Max Sum

Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
 
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
 
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
 
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
#include <bits/stdc++.h>
using namespace std;
const int MAXN = + ;
int T, n;
int arr[MAXN], dp[MAXN];
int S, E;
int main() {
scanf("%d", &T);
for(int t = ; t < T; ++t) {
S = E = ;
cin >> n;
for(int i = ; i != n; ++i)
cin >> arr[i];
dp[] = arr[];
for(int i = ; i != n; ++i)
if(dp[i-] >= )
dp[i] = dp[i-] + arr[i];
else
dp[i] = arr[i];
int Max = dp[];
for(int i = ; i != n; ++i)
if(dp[i] >= Max) {
Max = dp[i];
E = i;
}
int sum = ;
for(int i = E; i >= ; --i) {
sum += arr[i];
if(sum == Max)
S = i;
}
cout << "Case " << t+ << ":" << endl;
cout << Max << " " << S+ << " " << E+ << endl;
if(t < T-) puts("");
}
return ;
}

用了数组

#include <bits/stdc++.h>
using namespace std; int main() {
int T,n;
int Max, S, E, sum, a;
cin >> T;
for(int t = ; t <= T; ++t) {
cin >> n;
S = E = sum = ;
Max = -;
int k = ;
for(int i = ; i != n; ++i) {
cin >> a;
sum += a;
if(sum > Max) {
Max = sum;
S = k;
E = i;
}
if(sum < ) {
sum = ;
k = i + ;
}
}
cout << "Case " << t << ":" << endl;
cout << Max << " " << S+ << " " << E+ << endl;
if(t < T) puts("");
}
return ;
}

不用数组

HDU1003 简单DP的更多相关文章

  1. HDU 1087 简单dp,求递增子序列使和最大

    Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 ...

  2. Codeforces Round #260 (Div. 1) A. Boredom (简单dp)

    题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. ...

  3. codeforces Gym 100500H A. Potion of Immortality 简单DP

    Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/a ...

  4. 简单dp --- HDU1248寒冰王座

    题目链接 这道题也是简单dp里面的一种经典类型,递推式就是dp[i] = min(dp[i-150], dp[i-200], dp[i-350]) 代码如下: #include<iostream ...

  5. poj2385 简单DP

    J - 简单dp Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit ...

  6. hdu1087 简单DP

    I - 简单dp 例题扩展 Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:32768KB     ...

  7. poj 1157 LITTLE SHOP_简单dp

    题意:给你n种花,m个盆,花盆是有顺序的,每种花只能插一个花盘i,下一种花的只能插i<j的花盘,现在给出价值,求最大价值 简单dp #include <iostream> #incl ...

  8. hdu 2471 简单DP

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2571 简单dp, dp[n][m] +=(  dp[n-1][m],dp[n][m-1],d[i][k ...

  9. Codeforces 41D Pawn 简单dp

    题目链接:点击打开链接 给定n*m 的矩阵 常数k 以下一个n*m的矩阵,每一个位置由 0-9的一个整数表示 问: 从最后一行開始向上走到第一行使得路径上的和 % (k+1) == 0 每一个格子仅仅 ...

随机推荐

  1. ubuntu一些基本软件安装方法

    ubuntu一些基本软件安装方法 首先说明一下 ubuntu 的软件安装大概有几种方式:1. deb 包的安装方式deb 是 debian 系 Linux 的包管理方式, ubuntu 是属于 deb ...

  2. 【转】JavaWeb MVC

    -------------------------------------------------------------------------------------------------- 1 ...

  3. Mac OSX定位命令路径的方法

    可以使用which命令来定位一个命令. http://www.cyberciti.biz/faq/how-do-i-find-the-path-to-a-command-file/

  4. MyBatis源码分析(1)——整体依赖关系图

    后续补充更新

  5. python 模块基础介绍

    从逻辑上组织代码,将一些有联系,完成特定功能相关的代码组织在一起,这些自我包含并且有组织的代码片段就是模块,将其他模块中属性附加到你的模块的操作叫做导入. 那些一个或多个.py文件组成的代码集合就称为 ...

  6. js中的逻辑与(&&)和逻辑或(||)

    之前有一个同事去面试,面试过程中碰到这样一个问题: 在js中写出如下的答案 : var a = 2; var b = 3; var andflag = a && b ; var orf ...

  7. Codeforces Round #279 (Div. 2) ABCDE

    Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/outpu ...

  8. BZOJ4590——[Shoi2015]自动刷题机

    1.题意:题意很简洁吧,就不概括了 2.分析:我思考了半天,我猜答案满足单调...没敢写,看了题解去问Claris为啥单调,Claris一句话" 因为n越大明显不可能做更多题 ", ...

  9. php换行符

    1.需求 统一php换行符 2.实践 使用PHP_EOL替换换行符,保证平台的兼容性. 类似的有DIRECTORY_SEPARATOR 参考文档:http://www.cnblogs.com/code ...

  10. python pickle

    >>> import pickle >>> m_list=[',2,'asa'] >>> m_list [', 2, 'asa'] >> ...