【HDU】 1003 Max Sum（最大连续子序列和）

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

Ignatius.L

 

 

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AC代码：

#include<cstdio>int a[100000+10];int main(){    int t,n,k=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        int max=-1001,st,endd,sum=0,st1=1; //注意整数的范围        for(int i=1;i<=n;i++)             {            sum+=a[i];            if(sum>max)            {                max=sum;                st=st1;                endd=i;                }            if(sum<0)            {                sum=0;                st1=i+1;    //st1是临时的开始点，如果后面的sum<max，这个开始点也就不会计入            }            }        printf("Case %d:\n%d %d %d\n",k++,max,st,endd);        if(t!=0)        {            printf("\n");        }            }    return 0;}