HDU 1003 Max Sum

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

 

求最大子段和，基础题！

 

 1 #include <stdio.h>
 2 int main()
 3 {
 4     int t;
 5     int n;
 6     int a[100001];
 7     scanf("%d", &t);
 8     for(int i=1; i<=t; i++){
 9         scanf("%d",&n);
10         for(int j=1; j<=n; j++){
11             scanf("%d",&a[j]);
12         }
13         int p1,p2;
14         p1=p2=1;
15         int pos=1;
16         int temp=a[1];
17         int res=a[1];
18         for(int j=2; j<=n; j++){
19             //如果之前是负的话，就重新开始存
20             //-1 3 -1 2
21             //第一次是-1的时候就重新存
22             //第二次则不需要，因为之前的和为正数 
23             if(a[j]+temp<a[j]){
24                 temp=a[j];
25                 pos=j; 
26             }else{
27                 temp+=a[j];
28             }
29             //每次发现最大的时候则存放起来 
30             if(temp>=res){
31                 res=temp;
32                 p1=pos;
33                 p2=j;
34             }
35         }
36         if(i!=1)printf("\n");
37         printf("Case %d:\n", i);
38         printf("%d %d %d\n", res, p1, p2);
39     }
40     return 0;
41 }