148A - Insomnia cure

一位公主在夜晚想象所有龙都来偷窃,她通过每k、l、m、n个龙进行不同的惩罚,计算有多少只龙受到了道德或身体上的伤害。文章提供了一个输入数据格式,包括k、l、m、n和总龙数d,输出受到伤害的龙数量。这是一个关于整数计数和条件判断的数学问题。

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A. Insomnia cure
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.

However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every k-th dragon got punched in the face with a frying pan. Every l-th dragon got his tail shut into the balcony door. Every m-th dragon got his paws trampled with sharp heels. Finally, she threatened every n-th dragon to call her mom, and he withdrew in panic.

How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of d dragons?

Input

Input data contains integer numbers k, l, m, n and d, each number in a separate line (1 ≤ k, l, m, n ≤ 101 ≤ d ≤ 105).

Output

Output the number of damaged dragons.

Sample test(s)
input
1
2
3
4
12
output
12
input
2
3
4
5
24
output
17
Note

In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.

In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.

import java.util.*;

public class Main {
    public static void main(String[] args) {
        
        Scanner sc=new Scanner(System.in);
        int k=sc.nextInt();
        int l=sc.nextInt();
        int m=sc.nextInt();
        int n=sc.nextInt();
        int d=sc.nextInt();
        int[] a= new int[d+1];
        int count=0;
       solve(k,d,a);
       solve(l,d,a);
       solve(m,d,a);
       solve(n,d,a);
       for(int i=1;i<=d;i++)
           if(a[i]==1) count++;
        System.out.println(count);
    }
    public static void solve(int x,int d,int[] a)
    {
        int i=1;
        while(x*i<=d)
        {
            a[x*i]=1;
            i++;
        }
    }
}

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