HDU 1003 Max Sum（求序列最大子序列和）

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本文解析了 HDU 1003 Max Sum 的算法问题，介绍了如何通过动态规划的方法找到给定整数序列中具有最大和的子序列及其位置。提供了完整的 C++ 实现代码。

HDU 1003 Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 170306 Accepted Submission(s): 39737

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4 Case 2:7 1 6

【思路分析】

该题为求序列的最大序列和问题。

设d[i]表示序列a从1到i的最大序列和，则d[i] = max(d[i - 1] + a[i]，a[i])。

代码如下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int num[maxn];
void deriveSum(int n)
{
    int sum = 0,Max = num[1];
    int left = 1,right = 1,tmp = 1;
    for(int i = 1;i <= n;i++)
    {
        sum += num[i];
        if(sum > Max)
        {
            Max = sum;
            left = tmp;//更新左端点
            right = i;//更新右端点
        }
        if(sum < 0)
        {
            sum = 0;
            tmp = i + 1;
        }
    }
    printf("%d %d %d\n",Max,left,right);
}
int main()
{
    int t;
    scanf("%d",&t);

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

            printf("Case %d:\n",k);
            deriveSum(n);
            if(k < t)
                printf("\n");
        }

    return 0;
}