hdu 1003 Max sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 208607    Accepted Submission(s): 48835

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
 1 #include <iostream>
 2 #define N 100010
 3 using namespace std;
 4 int a[N], d[N];
 5 int main()
 6 {
 7     int test, n, i, max, k, b, e;
 8     cin >> test;
 9     k =1;
10 
11     while (test--)
12     {
13         cin >> n;
14         for ( i = 1; i <= n; i++)
15         {
16             cin >> a[i];
17         }
18         d[1] = a[1];
19         for ( i = 2; i <= n; i++)
20         {
21             d[i] = (d[i-1] + a[i] > a[i] ? d[i-1] + a[i] : a[i]);
22         }
23         max = d[1];
24         e = 1;
25         for (i = 2; i <= n; i++)
26         {
27             if(d[i] > max){
28                 max = d[i];
29                 e = i;
30             }
31         }
32         int t = 0;
33         b = e;
34         for (i = e; i > 0; i--)
35         {
36             t += a[i];
37             if (t == max){
38                 b = i;
39             }
40         }
41         cout << "Case " << k++ << ":" << endl << max << " " << b << " " << e << endl;
42         if(test)
43             cout << endl;
44     }
45     return 0;
46 }
View Code

 

posted @ 2016-05-10 08:43  琴影  阅读(159)  评论(0编辑  收藏  举报