http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 161361    Accepted Submission(s): 37794

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

代码：

 1 #include <fstream>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstdio>
 5 #include <cstring>
 6 #include <cmath>
 7 #include <cstdlib>
 8
 9 using namespace std;
10
11 #define EPS 1e-10
12 #define ll long long
13 #define INF 0x7fffffff
14
15 int main()
16 {
17     //freopen("D:\\input.in","r",stdin);
18     //freopen("D:\\output.out","w",stdout);
19     int T,n,ans,tn,l,r,al,ar,t;
20     scanf("%d",&T);
21     for(int tt=1;tt<=T;tt++){
22         scanf("%d",&n);
23         ans=tn=-INF;
24         for(int i=1;i<=n;i++){
25             scanf("%d",&t);
26             if(tn<0){
27                 l=r=i;
28                 tn=t;
29             }else{
30                 tn+=t;
31                 r=i;
32             }
33             if(tn>ans){
34                 al=l;
35                 ar=r;
36                 ans=tn;
37             }
38         }
39         printf("Case %d:\n%d %d %d\n",tt,ans,al,ar);
40         if(tt!=T)   puts("");
41     }
42     return 0;
43 }