最大子段和

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

 

 

题意：求最大的子段和，和它的起始位置和终止位置。

 

 

 

解题思路：

　　　　 在输入的时候，就可以开始求最大子段和了。当前子段和的值为负值时，就没必要再往下进行，而应将下一位数作为子段的第一位。也就是说，当处理第i个数时，如果以第i-1个数为结尾的子段的和为正数，则不必将第i个数作为新子段的首位进行枚举，因为新子段加上前面子段和所得的正数，一定能得到更大的子段和。然后在标记下位置就好了....

 

 

代码如下：

 

 

#include<stdio.h>
int a[100005];
int main()
{
    int T;
    int N=0;
    scanf("%d",&T);
    while(T--)
    {
        N++;
        int n,max=-1010,l=0,r=0,sum=0,temp=1;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
            if(sum>max)
            {
                max=sum;
                l=temp;
                r=i;
            }
            if(sum<0)
            {
                sum=0;
                temp=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",N,max,l,r);
        if(T)
            printf("\n");
    }
}