hdu 1003 Max Sum （最大连续子序列）

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#最大连续子序列

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本文详细介绍了如何通过动态规划解决MaxSum问题，即在一个整数序列中找到连续子序列的最大和。包括输入输出规范、算法实现及示例代码解析。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158643 Accepted Submission(s): 37111

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


#include<iostream>
#include<cstdio>
#include<cstring>
#define INF 0xfffffff
using namespace std;

int a[100005];
int dp[100005];

int main()
{
    int t;
    int n;
    int max;
    int s,ss,e;
    scanf("%d",&t);
    for(int k=1; k<=t; k++)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        max=-INF;
        s=e=ss=1;
        for(int i=1; i<=n; i++)
        {
            dp[i]=dp[i-1]+a[i];
            if(dp[i]<a[i])
            {
                dp[i]=a[i];
                ss=i;
            }
            if(dp[i]>max)
            {
                max=dp[i];
                s=ss;
                e=i;
            }
        }
        printf("Case %d:\n%d %d %d\n",k,max,s,e);
        if(k<t)
            printf("\n");
    }
    return 0;
}