Max Sum（HDU-1003）

最新推荐文章于 2022-11-06 15:03:56 发布

原创最新推荐文章于 2022-11-06 15:03:56 发布 · 979 阅读

0 ·

CC 4.0 BY-SA版权

HDU 同时被 2 个专栏收录

257 篇文章

订阅专栏

动态规划——线性 DP

69 篇文章

订阅专栏

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及其起始与结束位置的算法。通过动态规划的方法，高效地解决了该问题，并提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 6 6

Source Program

#include<cstdio>
#include<iostream>
#define N 100001
using namespace std;
int main()
{
    int t;
    int n;
    int num[N],dp[N];
    int max_num,sum;
    int pos_begin,pos_end;
    int k,i;

    scanf("%d",&t);
    for(k=1;k<=t;k++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);

        max_num=-1001;
        for(i=1;i<=n;i++)
        {
            dp[i]=max(num[i],dp[i-1]+num[i]);
            if(max_num<dp[i])
            {
                max_num=dp[i];//记录最大值
                pos_end=i;//记录最后位置
            }
        }

        sum=max_num;
        for(i=pos_end;i>0;i--)//从最后位置开始
        {
            sum-=num[i];//逐个递减
            if(sum==0)//减到0停止
            {
                pos_begin=i;//记录开始位置
                break;
            }
        }
        printf("Case %d:\n%d %d %d\n",k,max_num,pos_begin,pos_end);
        if(k<t)//格式控制
            printf("\n");
    }
    return 0;
}