Problem--148A--Codeforces--A. Insomnia cure

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小兔叽的甜甜圈

最新推荐文章于 2024-03-20 19:41:15 发布

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本文介绍了一个关于公主失眠计数的趣味算法问题，通过输入特定数值来计算遭受精神或身体伤害的“龙”的数量。文章提供了两种不同的实现方法，帮助读者理解如何通过编程解决这一问题。

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A. Insomnia cure
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 ≤ 10, 1 ≤ d ≤ 105).

Output
Output the number of damaged dragons.

Examples
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.

大致思路：输入a，b，c，d，e，输出1~e中，a，b，c，d倍数的个数
01

#include<stdio.h>
int main(){
    int a,b,c,d,e;
    scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);
    int num[e+1];
    for(int i=1;i<=e;i++)
    num[i]=1;
    for(int i=1;i<=e;i++){
        if(i%a==0||i%b==0||i%c==0||i%d==0)
        num[i]=0;
    }
    int count=0;
    for(int i=1;i<=e;i++){
        if(num[i]==0)
        count++;
    }
    printf("%d\n",count);
}

#include<stdio.h>
int main () {
    int ss=0,i,a,b,c,d,x;
    scanf("%d %d %d %d %d",&a,&b,&c,&d,&x);
    for(i=1;i<=x;i++)
        if(i%a==0 || i%b==0 || i%c==0 || i%d==0)
            ss++;
    printf("%d\n",ss);
    return 0;
}