hdu1003 Max Sum 简单规划-优快云博客

本文链接：https://blog.youkuaiyun.com/liu_tian_wei/article/details/74857668

本文解析了一道经典的编程题——最大子序列和问题，通过给出的示例代码详细介绍了如何寻找一组数中连续且和最大的子序列，并输出该子序列的和及其起始和结束位置。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 247988 Accepted Submission(s): 58604

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

题目大意就是给定一组数，找到连续的且和最大的一组数，输出和及开始结束的位置。

#include<iostream>
using namespace std;
int main()
{
    int i,j,k,t,n,m;
    int dp[100005];
    scanf("%d",&t);
    for(k=1;k<=t;k++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        scanf("%d",&dp[i]);
        int sum=0,maxn=-11111,first=1,last=0,flag=1;
        for(i=1;i<=n;i++)
        {
            sum+=dp[i];
            if(sum>maxn)
            {
                maxn=sum;
                first=flag;
                last=i;
            }
            if(sum<0)
            {
                sum=0;
                flag=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",k,maxn,first,last);
        if(k!=t)
        printf("\n");
    }
    return 0;
    
}