Hdu3415 Max Sum of Max-K-sub-sequence)

Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 4836 Accepted Submission (s): 1765

Problem Description
Given a circle sequence A [1], A [2], A [3] ...... A [n]. circle sequence means the left neighbor of A [1] is A [n], and the right neighbor 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 shoshould 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
-1-1-1-1-1-1-1

Sample Output
7 1 3
7 1 3
7 6 2
-1 1 1
Obtain the maximum value, start position, and end position when the sum of n consecutive numbers cannot exceed m (minus n if n is greater than) in the n number of composition rings)
I think it is still necessary to do this. Although it is a bit like the monotonous queue question previously done, I did this for the first time.

# Include <stdio. h> int a [100005], sum [200005], q [200005]; int main () {int t, n, m, I, j, maxx, e, s, first, last; scanf ("% d", & t); while (t --) {scanf ("% d", & n, & m ); sum [0] = 0; for (I = 1; I <= n; I ++) {scanf ("% d", & a [I]); sum [I] = sum [I-1] + a [I];} for (; I <= n + m; I ++) sum [I] = sum [I-1] + a [I-n]; // extend m backward to achieve the effect of loop first = last = 0; q [last ++] = 0; // this should be noted that WA has been used many times and does not have enough understanding of monotonous queues. maxx = a [1]; e = s = 1; for (I = 1; I <= n + m; I ++) {while (first <last & sum [Q [last-1]> sum [I-1]) // find the minimum value? Because sum [I]-sum [q [first], sum [I] is a fixed value // so to maximize this item, we must minimize sum [q [first] within the range of m {last --;} q [last ++] = I-1; // record is also I-1 while (first <last & q [first] <I-m) // remove first ++ out of range; if (maxx <sum [I]-sum [q [first]) {s = q [first] + 1 ;//, we recorded the I-1 at that time, here to calculate the starting point, we need to add 1 e = I; // The end point maxx = sum [I]-sum [q [first]; // maximum value} if (s> n) // do not drop s-= n; if (e> n) e-= n; printf ("% d \ n", maxx, s, e);} return 0 ;}