hdu 1003（最大连续子序列和）

 

　　　　Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 87019    Accepted Submission(s): 20165

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

/*说下起点和终点的问题如果子序列之和大于最大值的话那么初始化起点为j，如果子序列之和小于0，重新更新j，并且当子序列之和大于最大值，更新起点*/

#include<stdio.h>
#include<string.h>
#include<limits.h>
#include<algorithm>
using namespace std;
#define maxn 100001
int arr[maxn];
int main()
{
    int t,n,i,j;
    int st,nd,sub_max,max;
    scanf("%d",&t);
    int cnt;
    cnt=1;
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&arr[i]);
        sub_max=0;
        max=-1002;
        for(i=1,j=1;i<=n;i++)
        {
            sub_max+=arr[i];
            if(sub_max>max)
            {
                max=sub_max;
                st=j;
                nd=i;
            }
            if(sub_max<0)
            {
                sub_max=0;
                j=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",cnt++,max,st,nd);
        if(t!=0)
            printf("\n");
    }
    return 0;
}