【HDU】1003 Max Sum（动态规划、子序列最大和）

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最新推荐文章于 2022-07-13 08:30:00 发布

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分类专栏：动态规划自学编程 useful 文章标签： C语言

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本文介绍了一种求解整数序列中具有最大和的连续子序列的问题，并提供了一个使用动态规划方法的有效解决方案。该算法能够找出给定序列中最大和的子序列及其起始和结束位置。

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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

AC代码：

//
//  main.cpp
//  1003
//
//  Created by showlo on 2018/4/10.
//  Copyright © 2018年 showlo. All rights reserved.
//


#include <stdio.h>
int a[100100];

int main() {
    int N,n;
    int start,end,max,num;
    int temp;
    num=1;
    scanf("%d",&N);
    while (N--) {
        scanf("%d",&n);
        for (int i=0; i<n; i++) {
            scanf("%d",&a[i]);
        }
        max=a[0];
        start=0;
        end=0;
        temp=0;
        for (int i=1; i<n; i++) {   //动态规划
            if (a[i-1]>=0) {
                a[i]=a[i]+a[i-1];
            }
            else
                temp=i;
            if (a[i]>max) {
                max=a[i];
                end=i;
                start=temp;
            }
        }
        printf("Case %d:\n",num);
        printf("%d %d %d\n",max,start+1,end+1);
        if (N) {
            printf("\n");
        }
        num++;
    }
    return 0;
}