zoj1003-Max Sum (最大连续子序列之和)

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 }

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