Max sum。。。。。hdu oj----1003

Max sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 111859    Accepted Submission(s): 25831

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<stdio.h>
int main()
{
    int T,N,count=0,a[100010];
    scanf("%d",&T);
    while(T--)
    {
        count++;
        int start,s,end,e,max,sum,i;
        scanf("%d",&N);
        for(i=1;i<=N;i++)
        scanf("%d",&a[i]);
        /*此题主要意思是记录任意几个连续数字的和最大，并记录开始与结束的位置*/
        for(sum=max=a[1],s=e=start=end=1,i=2;i<=N;i++)
        {     /*开始与结束的位置是变化的*/
            if(sum<0){sum=0;start=i;end=i-1;}
            /*第一种情况，sum变为负数时，sum归零处理不会影响后面的判断，起始值变为i，末尾值变为i-1*/
            sum+=a[i]; end++;
            if(sum>max){s=start;e=end;max=sum;}
            /*第二种情况，s 记录当前的起始值，e 记录当前的末位置*/
        }
        printf("Case %d:\n",count);
        printf("%d %d %d\n",max,s,e);
        if(T>0) printf("\n");
    }
    return 0;
}