hdu 1003 简单的动态规划 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 109964    Accepted Submission(s): 25381

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

 

#include <iostream>
#include<stdio.h>
#include<math.h>
#define MAX 100000
#define MIN -99999999999
using namespace std;
int main()
{
    int T,n,a[100010],ma,s,r,l,f,i,k;
    scanf("%d",&T);
    for(k=1;k<=T;k++)
    {
        scanf("%d",&n);
        ma=MIN;
        s=0;
        l=r=f=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            s=s+a[i];
            if(s>ma)
            {
                ma=s;
                l=f;
                r=i;
            }
            if(s<0)
            {
                s=0;
                f=i+1;
            }
        }
        printf("Case %d:
",k);
        printf("%d %d %d
",ma,l+1,r+1);
        if(k!=T) printf("
");
    }
    return 0;
}