Didn't notice that problem is not for Java, so I solved it. Maybe someone will need the solution. Solution without checking of stacks for emptiness.


import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.StringTokenizer;

public class HelloWorld {

    public static void main (String[] args)     {
        HashMap<Integer,  ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
        Scanner reader = new Scanner(System.in);
        int n = reader.nextInt();
        reader.nextLine();
        for (int i = 0; i < n; i++) {
            String command = reader.nextLine();
            StringTokenizer tokenizer = new StringTokenizer(command);
            command = tokenizer.nextToken();
            if ("PUSH".equals(command)) {
                int key = Integer.parseInt(tokenizer.nextToken());
                int value = Integer.parseInt(tokenizer.nextToken());
                if (map.containsKey(key)) {
                    map.get(key).add(0, value);
                }
                else {
                    map.put(key, new ArrayList<Integer>());
                    map.get(key).add(0, value);
                }
            }
            else if ("POP".equals(command)) {
                int key = Integer.parseInt(tokenizer.nextToken());
                if (map.containsKey(key)) {
                    System.out.println(map.get(key).get(0));
                    map.get(key).remove(0);
                }
            }

        }
    }
}

Edited by author 31.10.2016 23:24

Edited by author 31.10.2016 23:26

Here is my not accepted solution:
#include<iostream>
#include<string>
using namespace std;

double P(int power[],string names[],int a,int n,int video_power,int on_or_off[],int w,int h){
a=video_power;
for(int i=0;i<n;i++){if(on_or_off[i]==1){a/=power[i];}}
double res=a/(w*h);
return res;
}

int main()
{

int n;
cin>>n;
string * names = new string[n];
int * power = new int[n];
for (int i=0;i<n;i++){
cin>>names[i];
cin>>power[i];
}

int w,h,video_power;
cin>>w>>h>>video_power;

double results[100];
int on_or_off[100];
for(int i=0;i<n;i++){on_or_off[i]=1;}
int e=0;

int a=video_power;
for (int i=0; i<n;i++){
a/=power[i];
}
double prodyctivity=a/(w*h);
results[e]=prodyctivity;e++;

int m;
cin>>m;
for(int i=0; i<m; i++){
string change,change_name;
cin>>change;
if(change[0]!='R'){
cin>>change_name;
for(int y=0;y<n;y++){
    if(change=="On" && change_name==names[y]){
        on_or_off[y]=1;
        prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
        results[e]=prodyctivity;e++;
        }
    else if(change=="Off" && change_name==names[y]){
        on_or_off[y]=0;
        prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
        results[e]=prodyctivity;e++;
        }
}
}
else{
int a1,a2;
cin>>a1>>a2;
w=a1;h=a2;
prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
results[e]=prodyctivity;e++;
}
}

for(int i=0;i<m+1;i++){
    if(results[i]<10){cout<<"Slideshow"<<endl;}
    else if(results[i]>=60){cout<<"Perfect"<<endl;}
    else{cout<<"So-so"<<endl;}
}
}

Please!

If the plane fly across a edge and it will produce a point then produce a new block so the answer add one.
If gcd(M, N) = 1, it means there is no situation that the plane fly across the crossroads.
See from left to right it produce M point and see from north to south it produce N point,
but one is calculate twice, so there are M + N - 1 points and Ans is M + N - 1.

I've perused over some of the other solutions here, tried to come up with something original, but I'm having a difficult time understanding why it won't stop letting me input in Eclipse, and how to deal with that...Here's my code:

import java.util.*;

public class ReverseRoot
{//start class
    public static void main(String[] args)
    {//start main
        Scanner in = new Scanner(System.in);
        ArrayList<Long> array = new ArrayList<Long>();
        array.add(in.nextLong());

        while(in.hasNextLong())
        {
            array.add(in.nextLong());
        }
        in.close();

        for (int i = array.size(); i > 0; i--)
            System.out.printf("%.4f%n", Math.sqrt((double)array.get(i)));
    }//end main
}//end class

s1 = ['1 2 3 4 5',
      '1 4 3 2 5',
      '1 3 2 4 5',
      '1 3 4 2 5']

think about :

F(N,M)  =  F(N-? , M- ? )

and give the initial values.

#include<stdio.h>
#include<string.h>
int f(int n,int m)
{
    if(n==1||m==1)
        return 0;
    if(n==2&&m>1)
        return 2;
    if(n==3&&m==2)
        return 3;
    if(m>2&&n>2)
    return f(n-2,m-2)+4;

}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    printf("%d",f(n,m));
    fflush(stdin);
    getchar();
    return 0;
}

* The minimal height of the wall is set to 1 (previously 0).
* The total number of bricks is set to be between 1 and 100 000 (previously, not more than 1 000 000, but there were no such tests).
* New tests have been added.
* All accepted solution have been rejudged, 98 authors have lost their AC.

[code deleted]

Edited by moderator 19.11.2019 23:13

It looks as first input number (sequence length) isn't needed for solution.

int a[10]={-1,1,2,2,1,-1,-2,-2,-1,1};
...
for(i=1;i<=8;i++) if(x+a[i-1]>0 && x+a[i-1]<9 && y+a[i+1]>0 && y+a[i+1]<9) h++;

DO IT!!! :)

Let C  = A + '$*&' + B
for string C build Palindrom Tree.
calculate frequency of each palindrom : f(A,p)
calculate frequency of each palindrom : f(B,p)
x = count(f(A,p)>f(B,p))
y = count(f(A,p)==f(B,p) && f(A,p)!=0)
z = count(f(A,p)<f(B,p))

//little code:  freq[i][0] ---> f(A,p);   freq[i][1] --> f(B,p).
for(int i = 0; i < a.size(); ++i){ add_letter(a[i]); freq[last][0]++;}

add_letter('$'); add_letter('*'); add_letter('&');

for(int i = 0; i < b.size(); ++i){ add_letter(b[i]); freq[last][1]++;}

for(int z = sz-1; z > 0; z--){
    v = link[z];
     freq[v][0] += freq[i][0];
     freq[v][1] += freq[i][1];
}
int x=0,y=0,z=0;
for(int i = 2; i< sz; ++i)// skip roots
{
    x += (bool)(freq[i][0] > freq[i][1]);
    y += (bool)(freq[i][0] == freq[i][1] && freq[i][0] !=0);
    z += (bool)(freq[i][0] < freq[i][1]);
}

#include <iostream>
#include <string>
using namespace std;
int main() {
    int num, res;
    string c, b = "!", a = "!";

    cin >> num >> c;
    res = num;
    int sign = 0;
    for (int i = 1; b <= c; i++) {
        b = b + a;
        sign = i;
    }
    if (num%sign != 0) {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
        res = res *(num%sign);
    }
    else {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
    }
    cout << res;

    return 0;
}

Solved both tasks with the same algorithm. Think this task is over-valued.

Edited by author 26.10.2016 20:59

just submitted code and got TLE on 15test, submitted same code after few minutes and got accepted. submitted again and got TLE on 24test for same code.

All solutions have been rejudged with the current time limit. Also, few more tests have been added. 10 authors have got AC while 23 other have lost.

RT

source code of algorithm Manaker (find all palindroms in O(n)), where explained in e-maxx.ru is incorrect :), see comments to below of that page.

Edited by author 25.10.2016 16:17