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[20000000],sum[200000000];
int main()
{
   int t,n,k,i,j,ij,m;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&k);
       m=n;
       a[0]=sum[0]=0;
       for(i=1,j=0;i<=n;i++,j++)scanf("%d",&a[i]),sum[i]=sum[j]+a[i];
       n<<=1;
       for(ij=1;i<=n;i++,j++,ij++)sum[i]=sum[j]+a[ij];
       int tail=0,top=0,maxx=sum[1],maxi=1,maxj=1;
       a[tail++]=0;
       n>>=1;
       n+=k;
       for(i=1;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]+1,maxj=i;
           while(tail>top&&sum[i]<sum[a[tail-1]])tail--;
           a[tail++]=i;
       }
       if(maxj>m)
       maxj-=m;
       printf("%d %d %d
",maxx,maxi,maxj);
   }
}