题目链接:https://ac.nowcoder.com/acm/contest/5667/F
解题思路:
单调队列详解:单调队列适用于解决一定长度的最大值最小值问题
#include<iostream>
#include<cstdio>
#include<vector>
#include<deque>
#include<string.h>
using namespace std;
#define ll long long
int n,m,p;
int a[5100][5100];
int b[5100][5100];
ll ans;
deque<int> q;
int main(){
cin>>n>>m>>p;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(!a[i][j]){
for(int k=1;k*i<=n&&k*j<=m;k++){
a[k*i][k*j]=k*i*j;
}
}
}
}
//对横向j进行单调队列操作
for(int i=1;i<=n;i++){
q.clear();
for(int j=1;j<=m;j++){
while(!q.empty()&&q.front()<=j-p) q.pop_front();
while(!q.empty()&&a[i][j]>=a[i][q.back()]) q.pop_back();
q.push_back(j);
if(j>=p)
b[i][j-p+1]=a[i][q.front()];
}
}
//纵向进行单调队列操作
for(int j=1;j<=m-p+1;j++){
q.clear();
for(int i=1;i<=n;i++){
while(!q.empty()&&q.front()<=i-p) q.pop_front();
while(!q.empty()&&b[i][j]>=b[q.back()][j]) q.pop_back();
q.push_back(i);
if(i>=p) ans+=b[q.front()][j];
}
}
cout<<ans<<endl;
return 0;
}