HDU 3415 Max Sum of Max-K-sub-sequence（求长度不超过K的最大区间和）

http://acm.hdu.edu.cn/showproblem.php?pid=3415

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

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经典单调队列问题，先将区间前缀和求出来，然后维护最小的单调队列。
每次判断的时候看是否在K长度范围之内即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int N = 1e5 + 100;const LL INF = (LL)N*N;LL f[N*2],que[N*2],arr[N*2];int pos[N*2];int main(){        int T;        scanf("%d",&T);        while(T--)        {                int n,k,x,idl,idr;                scanf("%d%d",&n,&k);                int ln = n;                for(int i=1;i<=n;i++) scanf("%I64d",&arr[i]),f[i]=f[i-1]+arr[i];                for(int i=1;i<k;i++) f[i+n] = f[i+n-1] + arr[i];                n += k-1;                LL ans = -INF;                  int head = 1, tail = 0;                for(int i=0;i<=n;i++)                {                        while(head <= tail && pos[head]+k<i) head++;                        if( head<=tail && ans < f[i]-que[head])                                 ans = f[i]-que[head],idl = pos[head]+1,idr = i;                        while(head <= tail && que[tail]>f[i]) tail--;                        que[++tail] = f[i];                        pos[tail] = i;                }                if(idl>ln)  idl -= ln;                if(idr>ln)  idr -= ln;                printf("%I64d %d %d\n",ans,idl,idr);        }        return 0;}