HDU 1003 Max Sum

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本文解析了一个经典的编程问题——最大子数组和问题，并提供了一种有效的解决方案。通过对输入序列的遍历，算法能够找到具有最大和的子数组及其位置，适用于算法竞赛和技术面试等场景。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 247744 Accepted Submission(s): 58531

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

分析：经典的最大字段和+记录路径理解了转换公式就ok了；

详见代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int n,t;
    int a[100005];
    int sum,MAX;
    int start,end,tem;
    int index=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n); 
        MAX=-100000000;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sum=start=tem=0;
        for(int i=0;i<n;i++)
        {
            if(sum>0)	//可能有的同学会纠结这个代码连实示例都不过，改为sum>=0就行了，其实两种方法都能找到最优解，只是区间记录的不一样 
                sum+=a[i];
            else
            {
                tem=i;
                sum=a[i];
            }
            if(sum>MAX)
            {
                MAX=sum;
                start=tem;	//这里必须借助tem变量，记录随时变化的左区间 
                end=i;
            }
        }
        printf("Case %d:\n",++index);
        printf("%d %d %d\n",MAX,start+1,end+1);
        if(t>0)
            printf("\n");
    }
    return 0;
}