文章目录
一、Insomnia cure
总结
一、Insomnia cure
本题链接：Insomnia cure
题目：
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.
本博客给出本题截图：
题意：首先输入四个数，然后输入一个数d，问在1~d中有多少个数是这四个数的倍数
AC代码
#include <iostream> using namespace std; const int N = 10; int a[N]; int main() { for (int i = 0; i < 4; i ++ ) cin >> a[i]; int d; cin >> d; int res = 0; for (int i = 1; i <= d; i ++ ) for (int j = 0; j < 4; j ++ ) { if (i % a[j] == 0) break; if (i % a[j] != 0 && j == 3) res ++; } cout << d - res << endl; return 0; }
总结
水题，不解释
