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Max sum of max-K-Sub-Sequence

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Problem descriptiongiven 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. inputthe 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 ). outputfor 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 
46 36 -1 2 -6 5 -56 46 -1 2 -6 5 -56 3-1 2 -6 5 -5 66 6-1 -1 -1 -1 -1 -1
Sample output 
7 1 37 1 37 6 2-1 1 1
Authornotonlysuccess @ hdusourcehdoj monthly contest-2010.06.05 struggled with the question of the day and thought wrong at first .. I thought it was just done .. Later I figured it out that it was wrong, and then I used the monotonous queue to solve it perfectly! Ans = max (sum [I]-min (sum [I-K]) adopts monotonous queue maintenance in the interval [I-K, the sum value of I-1] is monotonous, And the monotonous queue is very powerful! # Include <iostream> # include <algorithm> using namespace STD; # define n 200020 # define INF 0 xfffffffstruct node {int index, num;}; node Dia [N]; int Q [2 * n], sum [N]; int main () {int I, j, T, Head, tail; int N, K; scanf ("% d", & T); While (t --) {int S, E, ANS =-INF; sum [0] = 0; scanf ("% d", & N, & K); for (I = 1; I <= N; I ++) {scanf ("% d ", & Dia [I]. num); If (ANS <Dia [I]. num) {ans = Dia [I]. num; S = I; E = I;} Dia [I]. index = I; sum [I] = sum [I-1] + Dia [I]. num;} for (I = 1; I <K; I ++) {Dia [I + N]. index = I + N; Dia [I + N]. num = Dia [I]. num; sum [I + N] = sum [I + n-1] + Dia [I + N]. num;} Dia [0]. index = 0; Dia [0]. num = 0; head = N; tail = N; Q [+ head] = 0; int temp; for (I = 1; I <= N + k-1; I ++) {temp = sum [I]; if (I <= K & Ans <temp) {ans = temp; S = 1; E = I ;} while (DIA [Q [head]. index <I-k) head --; temp-= sum [Q [head]; If (ANS <temp) {ans = temp; S = Q [head] + 1; E = I ;} while (tail <Head & (sum [Q [tail + 1]> sum [I] | Q [tail + 1] <I-k) tail ++; Q [tail --] = I;} cout <ans <"" <S <"<(E> n? E-N: E) <Endl;} return 0 ;}

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