最大子序列及标记首尾坐标

Max Sum 
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) 
Total Submission(s): 52982    Accepted Submission(s): 11890 


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

算法：

#include<cstdio>#include<cstring>#include<map>#include<vector>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint main(){int T;int n;int Case = 1; int num[100000+5];scanf ("%d",&T);while (T--){scanf ("%d",&n);for (int i = 1 ; i <= n ; i++)scanf ("%d",&num[i]);int sum = 0;int ans = -INF;int tst = 1;//临时存放起点int endd = tst;int st = 1;// 5 3 -5 -4 1 2 -5for (int i = 1 ; i <= n ; i++){sum += num[i];if (sum > ans){ans = sum;st = tst;endd = i;}if (sum < 0){sum = 0;tst = i+1;}}printf ("Case %d:\n",Case++);printf ("%d %d %d\n",ans,st,endd);if (T != 0)printf ("\n");}return 0;}