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.

C:

#include <stdio.h>  
int main()  
{  
    int z,n,max,sum;  
    int a,b,A,B,t;  
    scanf("%d",&z);  
    for(int k=1;k<=z;k++)  
    {  
        scanf("%d",&n);  
        sum = max = -1001;  
        for(int i=1;i<=n;i++)  
        {  
            scanf("%d",&t);  
            if(sum+t < t)  
                sum = t , a = b = i;      //a、b记录当前连续子序列的起始、结束位置  
            else  
                sum += t , ++b;  
            if(max < sum)  
                max = sum , A = a , B = b;  
        }  
        printf("Case %d:\n%d %d %d\n",k,max,A,B);  
        if(k-z) puts("");  
    }  
    return 0;  
}  

C++:

#include <iostream>
#include <cstdio>

using namespace std;
const int inf = 0x7fffffff;

int main()
{
   int n, num, t, tt=1;
   bool first = true;

   scanf("%d", &t);
   while(t--)
   {
       scanf("%d", &n);
       int maxsum = -inf;
       int ans_s = 0, ans_e = 0;
       int cur = 1, sum = 0;

       for(int i=1; i<=n; i++)
       {
           scanf("%d", &num);
           sum += num;
           if(sum>maxsum)
           {
               maxsum = sum;
               ans_s = cur;
               ans_e = i;
           }
           if(sum<=0)
           {
               sum = 0;
               cur = i+1;
           }
       }
       if(first)
       {
           first = false;
           printf("Case %d:\n%d %d %d\n", tt++, maxsum, ans_s, ans_e);
       }
       else
       {
           printf("\nCase %d:\n%d %d %d\n", tt++, maxsum, ans_s, ans_e);
       }
   }
   return 0;
}

Java:

import java.util.Scanner;

public class Main {

    public static void main(String args[]){
        Scanner s=new Scanner(System.in);
        int T=s.nextInt();
        for(int i=0;i<T;i++){
            int n=s.nextInt();
            int sum=0,max=-1001;
            int start=0,end=0,z=0;
            for(int j=0;j<n;j++){
                int a=s.nextInt();
                sum=sum+a;
                if(max<sum){
                    max=sum;
                    end=j;
                    start=z;
                }
                if(sum<0){
                    sum=0;
                    z=j+1;
                }
            }
            System.out.println("Case "+(i+1)+":");
            System.out.println(max+" "+(start+1)+" "+(end+1));
            if(i<T-1)
                System.out.println();
        }
    }
}