Max Sum

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Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98077    Accepted Submission(s): 22598

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
   

 

Author

据说求最大字段和要用dp做，我没学过dp，不知道这种方法是不是，反正此方法，wrong了n+1次，才通过，真是黄天不负有心人啊。

#include<stdio.h>
int z[100007];
int main()
{
    int m,count=0,w,sum,max;
    int n,start,end,p;
    scanf("%d",&m);
    w=m;
    while(m--)
    {

        scanf("%d",&n);
        for(int c=1;c<=n;c++)
        scanf("%d",&z[c]);
         sum=z[1],max=z[1];
         p=1,start=1,end=1;
        for(int b=2;b<=n;b++)
        {
            if(sum<0)
            {
                p=b;
                sum=0;
            }
            sum+=z[b];
            if(sum>max)
            {
                end=b;
                start=p;
                max=sum;
            }
        }
        count++;
        if(count<w)
        {
            printf("Case %d:\n",count);
        printf("%d %d %d\n\n",max,start,end);
        }
        else
        {
            printf("Case %d:\n",count);
        printf("%d %d %d\n",max,start,end);
        }

    }
    return 0;
}