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.
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).
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.
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> #include <algorithm> #include <cmath> using namespace std; const int MAXN = 100010; int N, K; int a[MAXN]; int sum[2 * MAXN]; /* * 由于是统计不超过K个连续元素和的最大值，我们以sum值作为处理对象。 * 思想：线性枚举当前状态，并且维护在K范围内的最小值，对就是用单调 * 队列取维护当前允许范围内的最小值，然后求出当前最有解去更新总体的最优解即可 * 很简单的，实际上就应该是传说中的斜率优化。理解万岁，为什么当年就是想不出来呢？ */ struct node { int num, index; } que[2 * MAXN]; int front, tail; void init() { front = 0, tail = -1; } void Insert(int pos, int num) { if (front > tail) { front = tail = 0; que[front].index = pos; que[front].num = num; return; } while (front <= tail && que[tail].num >= num) tail--; que[++tail].index = pos; que[tail].num = num; while (front <= tail && pos - que[front].index + 1 > K) front++; } int main() { int T; scanf("%d", &T); while (T--) { scanf("%d%d", &N, &K); init(); sum[0] = 0; for (int i = 1; i <= N; i++) { scanf("%d", &a[i]); sum[i] = sum[i - 1] + a[i]; } for (int i = 1; i <= N; i++) { sum[i + N] = sum[i + N - 1] + a[i]; } int l, r; int ans = -100000000; Insert(0, sum[0]); for (int i = 1; i <= 2 * N; i++) { if (sum[i] - que[front].num > ans) { ans = sum[i] - que[front].num; l = que[front].index + 1; r = i; } Insert(i, sum[i]); } if (l > N) l -= N; if (r > N) r -= N; printf("%d %d %d\n", ans, l, r); } return 0; }