Submission #304217

#	Time	Username	Problem	Language	Result	Execution time	Memory
304217		KWang31	Fancy Fence (CEOI20_fancyfence)	Java	0 / 100	173 ms	13312 KiB

fancyfence

import java.io.*; import java.util.*;
public class fancyfence {

    /**
     * Tested on https://codeforces.com/problemset/problem/1402/A
     * See analysis
     */
    public static class Pair{
        int h; long len;
        public Pair(int a, long b){
            this.h=a; this.len=b;
        }
    }
    static int MOD=1000000007;
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int N=Integer.parseInt(br.readLine());
        int[] h=new int[N]; int[] w=new int[N];
        
        StringTokenizer st=new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            h[i]=Integer.parseInt(st.nextToken());
        }
        st=new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            w[i]=Integer.parseInt(st.nextToken());
        }
        
        
        long ans=0;
        Stack<Pair> stk=new Stack<>();
        long cur; long x,y,cur2,z; 
        for (int i = 0; i < N; i++) {
            //Only count rectangles whose RIGHT SIDE lies in rectangle i
            
            //Stack is monotonic: Pop everyone greater than h, add correspondingly
            cur=0;//Tracks the width of stuff with height >h
            while(!stk.isEmpty() &&stk.peek().h>h[i]){
                cur+=stk.pop().len;
            }
            
            x=(long) h[i]*(h[i]+1)/2; x%=MOD;
            cur*=x; cur%=MOD;
            cur*=w[i]; cur%=MOD;
            
            ans+=cur; ans%=MOD;
            
            //Now count the rectangles in the current rectangle
            y=(long) w[i]*(w[i]+1)/2; y%=MOD;
            ans+=(x*y)%MOD; ans%=MOD;
            //Now (naively) count remaining rectangles in the stack, this can be sped up w/ prefix sum
            
            if(!stk.isEmpty()){
                //Merge sections
                //For everyone in the stack, add some stuff
                cur2=0;
            
                for (Pair q: stk) {
                    z=(long) q.h*(q.h+1)/2; z%=MOD;
                    cur2+=q.len*z; cur2%=MOD;
                }
                cur2*=w[i];
                ans+=cur2;
                Pair p=stk.pop();
                stk.add(new Pair(p.h,p.len+cur));
            }
            
            stk.add(new Pair(h[i],w[i]));
            //System.out.println(ans);
        }
        System.out.println(ans);
    }
    
}

• Subtask 1: 0 / 0
• Subtask 2: 0 / 12
• Subtask 3: 0 / 13
• Subtask 4: 0 / 15
• Subtask 5: 0 / 15
• Subtask 6: 0 / 18
• Subtask 7: 0 / 27