Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4913 Accepted Submission(s): 1791
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 <cstdio>
#include <cstring>
using namespace std;
int a[222222],sum[222222]={0},que[222222];
int main (void)
{
int t,n,m,s,i,j,k,l,max,aa,ss,qian,hou;
scanf("%d",&t);
while(t--&&scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
scanf("%d",&a[i]),a[n+i]=a[i];
for(i=1,sum[0]=0;i<=n+m;i++)
sum[i]=sum[i-1]+a[i]; //把總和記錄下來
max=-1000000,aa=ss=qian=hou=0;
for(i=1;i<n+m;i++)
{
while(qian<hou&&sum[i-1]<sum[que[hou-1]])hou--; //保持單調
que[hou++]=i-1;
while(qian<hou&&i-que[qian]>m)qian++; //保持長度
if(sum[i]-sum[que[qian]]>max)
{
max=sum[i]-sum[que[qian]];
aa=que[qian]+1;ss=i;
}
}
if(aa>n)aa-=n;
if(ss>n)ss-=n;
printf("%d %d %d\n",max,aa,ss);
}
return 0;
