huu 1003Max Sum dp

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

思路：在循环的过程中，每循环一次就算出一个以当前位置结束的最大子序列和。每次循环中最大的那个保存下来，就是最终所有最大子序列和中的最大值

#define _CRT_SBCURE_MO_DEPRECATE#include<iostream>#include<stdlib.h>#include<stdio.h>#include<cmath>#include<algorithm>#include<string>#include<string.h>#include<set>#include<queue>#include<stack>#include<functional> using namespace std;int n, t;int a[100050];int sum;int str, en, k;int maxs;int main(){cin >> n;int temp = 1;int temp2 = n;while (n--) {cin >> t;sum = 0;maxs = -99999;str = en = k = 1;for (int i = 1; i <= t; i++)  cin >> a[i];for (int i = 1; i <= t; i++) {sum = sum + a[i];if (sum > maxs) { //现在的序列和比之前的最大子序列大；maxs = sum;//cout << maxl ;str = k;//开始位置就是上次序列和清零后重新开始新序列的地址en = i;//不管下一个加正还是负数，都会使下一个数增加；//cout << " " << en << endl;}if (sum <0) { //子序列和小于0，那么之后不管再加什么数，都会使下一个数减小；k = i + 1;//保存的是后一个位置的子序列和的开始位置sum = 0; // 序列和清零，从下一循环开始从新计算序列和}}printf("Case %d:\n%d %d %d\n", temp, maxs, str, en);if (temp != temp2)cout << endl;temp++;}//system("pause");return 0;}