至于题是哪来的，老师BB出来的
至于怎么BB的，请自己联想
Max Sum of Max-K-sub-sequence
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
题意自己百度
这道题，可以这样做
解释都在注释和下面
`
由于是一个圈
1 2 3 4 5
显然1的左边还有5
那么我们就把问题分为两类
第一类：
就是一个没有跨过了n的序列
第二类：
一个跨过了n的序列`
第一类
题意可转换为
有一序列
在其中找一个长度不超过k的子序列使其和最大
有多个子序列满足条件的按题意输出
就用。。。
第二类
在整个序列中找一子序列，使其总和最小且长度大于等于n - k
ans = ALL - 子序列总和
是不是很简单( ⊙ o ⊙ )啊！
#include <cstdio>
#include <climits>
const int MAXN = 100001;
class MYOWN
{
public:
int val,it;
MYOWN()
{
val = 0;
it = 0;
}
};
int main()
{
int T;
scanf("%d",&T);
//多组数据
while(T--)
{
int i,n,k,head,tail,ALL = 0;
/*
显然head记录队头，tail记录队尾
ALL方便第二种
*/
int num[MAXN] = {},sum[MAXN] = {};
int ans = -1001,l,r;
MYOWN q[MAXN];
scanf("%d%d",&n,&k);
for(i = 1;i <= n;i++)
{
scanf("%d",&num[i]);
sum[i] = sum[i - 1] + num[i];
ALL += num[i];
}
head = 1;tail = 1;
for(i = 1;i <= n;i++)
{
while(q[head].it < i - k&&head <= tail)
head++;
int delta = sum[i] - q[head].val;
if(delta > ans)
{
ans = delta;
l = q[head].it + 1;
r = i;
}
while(sum[i] <= q[tail].val&&tail >= head)
tail--;
tail++;
q[tail].val = sum[i];
q[tail].it = i;
}
//第一类，单调队列嘛
k = n - k;
q[0].val = -1001;
for(i = 1;i <= n;i++)
{
if(sum[i] <= q[i - 1].val)
{
q[i].it = q[i - 1].it;
q[i].val = q[i - 1].val;
} else {
q[i].val = sum[i];
q[i].it = i;
}
}
//第二类，根本不需要单调队列！！用用动规思想记录下
//q[i].val 记录从一到i的sum[i]的最大值
//q[i].it记录q[i].val第一次出现的下标（题目要求的多个答案的输出要求）
for(i = k + 1;i < n;i++)
{
int delta = ALL - sum[i] + q[q[i - k].it].val;
if(delta > ans)
{
ans = delta;
l = i + 1;
r = q[i - k].it;
}
}
printf("%d %d %d\n",ans,l,r);
}
return 0;
}