Max Sum of Max-K-sub-sequence hdu3415

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5569    Accepted Submission(s): 2003

Problem Description

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output a 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 minimum start position, if still more than one , output the minimum length of them.

Sample Input

4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1

Sample Output

7 1 3
7 1 3
7 6 2
-1 1 1

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 int a[],sum[];
 int main()
 {
    int t,n,k,i,j,ij,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        m=n;
        a[]=sum[]=;
        for(i=,j=;i<=n;i++,j++)scanf("%d",&a[i]),sum[i]=sum[j]+a[i];
        n<<=;
        for(ij=;i<=n;i++,j++,ij++)sum[i]=sum[j]+a[ij];
        int tail=,top=,maxx=sum[],maxi=,maxj=;
        a[tail++]=;
        n>>=;
        n+=k;
        for(i=;i<n;i++)
        {
            while(tail>top&&i-a[top]>k)top++;
            if(maxx<sum[i]-sum[a[top]])
                maxx=sum[i]-sum[a[top]],maxi=a[top]+,maxj=i;
            while(tail>top&&sum[i]<sum[a[tail-]])tail--;
            a[tail++]=i;
        }
        if(maxj>m)
        maxj-=m;
        printf("%d %d %d\n",maxx,maxi,maxj);
    }
 }