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：

#include <fstream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>

using namespace std;

#define EPS 1e-10
#define ll long long
#define INF 0x7fffffff

int main()
{
    //freopen("D:\input.in","r",stdin);
    //freopen("D:\output.out","w",stdout);
    int T,n,ans,tn,l,r,al,ar,t;
    scanf("%d",&T);
    for(int tt=1;tt<=T;tt++){
        scanf("%d",&n);
        ans=tn=-INF;
        for(int i=1;i<=n;i++){
            scanf("%d",&t);
            if(tn<0){
                l=r=i;
                tn=t;
            }else{
                tn+=t;
                r=i;
            }
            if(tn>ans){
                al=l;
                ar=r;
                ans=tn;
            }
        }
        printf("Case %d:
%d %d %d
",tt,ans,al,ar);
        if(tt!=T)   puts("");
    }
    return 0;
}