HDU——1003Max Sum（子序列最大和）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 197869    Accepted Submission(s): 46229

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

 

 

对于一个数，决策只有两种，要么让他跟前面的并在一起，要么前面的扔掉，从这个数开始取。

用dp结构体来表示，比较直观

若把前面的一段与A[i]合并，则当前点的l为dp[i-1].l保持不变，r=dp[i-1].r+1，因为此时A[i]算了进去；

若直接从A[i]，那么令dp[i].l=dp[i].r=i即可，dp[i].val=A[i]。

最后找到一个最大的dp[i].val即可

代码：

#include <stdio.h>
const int N = 100010;
struct info
{
    int val;
    int l, r;
};
info dp[N];
int arr[N];


int main(void)
{
    int tcase, n, i;
    scanf("%d", &tcase);
    for (int q = 1; q <= tcase; ++q)
    {
        scanf("%d", &n);
        for (i = 1; i <= n; ++i)
            scanf("%d", arr + i);
        dp[1].val = arr[1];
        dp[1].l = 1;
        dp[1].r = 1;
        for (i = 2; i <= n; ++i)
        {
            int a = dp[i - 1].val + arr[i];
            int b = arr[i];
            if (a >= b)
            {
                dp[i].val = a;
                dp[i].l = dp[i - 1].l;
                dp[i].r = dp[i - 1].r + 1;
            }
            else
            {
                dp[i].l = dp[i].r = i;
                dp[i].val = arr[i];
            }
        }
        int indx = 1;
        for (i = 1; i <= n; ++i)
            if (dp[i].val > dp[indx].val)
                indx = i;
        printf("Case %d:\n%d %d %d\n%s", q, dp[indx].val, dp[indx].l, dp[indx].r, q != tcase ? "\n" : "");
    }
    return 0;
}