HDU 1003 Max Sum

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本文详细解析了一种用于寻找给定整数序列中最大和子序列的算法。通过实例说明了算法的工作原理，介绍了输入输出格式，并提供了一个C++实现代码示例，帮助读者理解和实现该算法。

题目：

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

思路：

将数组累加，在累加的过程中判断，如果大于ans的话，则更新ans和头指针和尾指针，如果累加和小于0，则将累加和置为0，临时头指针置为i+1。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int t;
int n;
int a[maxn];
int main()
{
    scanf("%d",&t);
    for (int j=1;j<=t;j++)
    {
        scanf("%d",&n);
        for (int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int st=0,tst=0,en=0,ans=a[0],temp=0;
        for (int i=0;i<n;i++)
        {
            temp+=a[i];
            if(temp>ans)
            {
                st=tst;
                ans=temp;
                en=i;
            }
            if(temp<0)
            {
                tst=i+1;
                temp=0;
            }
        }
        printf("Case %d:\n",j);
        printf("%d %d %d\n",ans,st+1,en+1);
        if(j!=t)
            printf("\n");
    }
    return 0;
}