Title Description
Comprising a number of columns (n <= 2000000) n items, obtains the number m of each of the front to a minimum in this range it. When the number is less than m in front of the number of items from a first start, if there is no front number 0 is output.
Input Format
The first line of the two numbers n, m.
Second row, n-positive integer, for the given number of columns.
Output Format
n rows, a number ai i-th row, to the minimum required number of m before the i-th sequence.
Sample input and output
Input # 1
6 2 7 8 1 4 3 2
Output # 1
0 7 7 1 1 3
Description / Tips
[Data] scale
m≤n≤2000000
to i ≤ 3 × 1 0 ^ 7
analysis
Monotonous queue
Internal queues monotonically increasing monotone
Code
1 /************************** 2 User:Mandy.H.Y 3 Language:c++ 4 Problem:luogu1440 5 Algorithm: 6 **************************/ 7 #include<bits/stdc++.h> 8 9 using namespace std; 10 11 const int maxn = 2e6 + 5; 12 13 int n,l,r,m; 14 int q[maxn],a[maxn]; 15 16 template<class T>inline void read(T &x){ 17 x = 0;bool flag = 0;char ch = getchar(); 18 while(!isdigit(ch)) flag |= ch == '-',ch = getchar(); 19 while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar(); 20 if(flag) x = -x; 21 } 22 23 template<class T>void putch(const T x){ 24 if(x > 9) putch(x / 10); 25 putchar(x % 10 | 48); 26 } 27 28 template<class T>void put(const T x){ 29 if(x < 0) putchar('-'),putch(-x); 30 else putch(x); 31 } 32 33 void file(){ 34 freopen("1440.in","r",stdin); 35 freopen("1440.out","w",stdout); 36 } 37 38 void readdata(){ 39 read(n);read(m); 40 } 41 42 void work(){ 43 for(int i = 1;i <= n; ++ i){ 44 45 read(a[i]); 46 47 while(l < r && i - q[l] > m) l++; 48 49 if(l >= r) puts("0"); 50 else put(a[q[l]]),putchar('\n'); 51 52 while(l < r && a[q[r - 1]] >= a[i]) r--;//>= 53 q[r++] = i; 54 } 55 } 56 57 int main(){ 58 // file(); 59 readdata(); 60 work(); 61 return 0; 62 }