HDU 1003 - Max Sum（难度：*）

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

 

【题意】 给n个数，求最大连续元素的和；并输出起点和终点。

【分析】

　　设d[i]为以第i个元素为终点的最大连续元素和。则

　　状态转移方程：d[i] = d[i-1] > 0 ? d[i-1]+a[i] : a[i];

 

　　思路应该比较好理解，如果d[i-1]<0， 那么无论a[i]为何值，其和总不如单独的a[i]大；反之如果d[i-1]>0， 那么无论a[i]为何值，其和总大于单独的a[i]；

【代码】　　

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 using namespace std;

 6 const int maxn = 100010;

 7 int n, a[maxn];

 8 int d[maxn];

 9 void dp()

10 {

11     memset(d, 0, sizeof(d));

12     d[0] = a[0];

13     for(int i = 1; i < n; i++)

14     {

15         if(d[i-1] > 0) d[i] = d[i-1]+a[i];

16         else d[i] = a[i];

17     }

18     //cout << endl;

19     int st = 0, en = 0;

20     int max_ = d[0];

21     for(int i = 1; i < n; i++)

22     {

23         if(d[i]>max_)

24         {

25             max_ = d[i];

26             en = i;

27         }

28     }

29     st = en;

30     for(int j = en-1; j>=0; j--)

31     {

32         if(d[j] < 0) break;

33         else st = j;

34     }

35     printf("%d %d %d\n", max_, st+1, en+1);

36 }

37 

38 int main()

39 {

40     int T; scanf("%d", &T);

41     for(int kase = 0; kase < T; kase++)

42     {

43         scanf("%d", &n);

44         for(int i = 0; i < n; i++)

45             scanf("%d", &a[i]);

46         if(kase) printf("\n");

47         printf("Case %d:\n", kase+1);

48         dp();

49 

50     }

51     return 0;

52 }

View Code