Max Sum HDU - 1003 求最大和

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本文详细解析了HDU1003题目的解决方案，介绍了如何通过编程算法找出给定整数序列中具有最大和的子序列及其起始与结束位置。

HDU 1003

题目描述

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

AC CODE :

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int main(void) {
    int t, times = 0;
    cin >> t;
    while (t--) {
        printf("Case %d:\n", ++times);
        int n, ni[100100];
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> ni[i];
        }
        
        int sum = -0x3f3f3f3f, maxn = -0x3f3f3f3f;
        int A, B, a = 0, b = 0;
        for (int i = 0; i < n; i++) {
            if (sum + ni[i] < ni[i]) {
            sum = ni[i]; a = b = i;
            } else {
                sum += ni[i]; b++;
            } 
            if (sum > maxn) {
                maxn = sum, A = a+1, B = b+1;
            }
        }
        printf("%d %d %d\n", maxn, A, B);
        if (t) puts("");
    }
    return 0;
}