Max Sum

                        Max Sum

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

 解析：最大子段和

 给定一个长度为n的一维数组a，找出此数组的一个子数组,使得此子数组的和sum＝a[i]+a[i+1]+……+a[j]最大，其中i>=0,i<n,j>=i,j<n

 设dp[i]表示以i为结尾的子段和，则状态转移方程为：

 dp[i] = max{dp[i-1]+a[i]，a[i]}    i>=1

# include<stdio.h>
struct node
{
    int st;//开始位置
    int en;//结束位置
    int sum;
    node()
    {
        st=1;
        en=1;
        sum=0;
    }
}dp[100002];
int main()
{
    int nCase,nNum;
    int i,j;
    int ma,mx,sign;
    scanf("%d",&nCase);
    for(i=1;i<=nCase;i++)
    {
        scanf("%d",&nNum);
        ma=-99999;
        mx=1;
        dp[0].sum=-1;
        for(j=1;j<=nNum;j++)
        {
            scanf("%d",&sign);
            if(sign>dp[j-1].sum+sign)//记录的一定是以sign结尾的最大和
            {
                dp[j].sum=sign;
                dp[j].st=j;
                dp[j].en=j;
            }
            else 
            {
                dp[j].sum=dp[j-1].sum+sign;
                dp[j].st=dp[j-1].st;
                dp[j].en=j;
            }
            if(ma<dp[j].sum)
            {
                ma=dp[j].sum;
                mx=j;
            }
        }
        printf("Case %d:\n%d %d %d\n",i,ma,dp[mx].st,dp[mx].en);
        if(i!=nCase)printf("\n");
    }
    return 0;
}