hdu---1003---Max Sum

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 216737    Accepted Submission(s): 51087

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

 

Author

Ignatius.L

 

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include<vector>
#include<algorithm>
using namespace std;
#define PI 3.1415926

const int maxn=1000007;
const int INF=0x3f3f3f3f;

int a[maxn];
int main()
{
    int T, cas=1, f=0;
    scanf("%d", &T);
    while(T--)
    {
         if(f)
            puts("");///printf("
");
        int n;
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
            scanf("%d", &a[i]);

        int nows=1, Start=1, End=1;
        int sum, Max;
        sum=Max=a[1];

        for(int i=2; i<=n; i++)
        {
            if(sum+a[i]<a[i])
            {
                sum=a[i];
                nows=i;
            }
            else
                sum+=a[i];

            if(sum>Max)
            {
                Max=sum;
                Start=nows;
                End=i;
            }
        }
        printf("Case %d:
%d %d %d
", cas++, Max, Start, End);
        f=1;
    }
    return 0;
}