NYOJ--How many integers can you find
2014-04-09 15:38
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How many integers can you find
时间限制:1000 ms | 内存限制:65535 KB难度:1
描述给你三个数,n,m1,m2,找出所有小于n的能被m1或m2整除的数的个数。
输入输入包含多组测试数据,每组数据占一行。
0<n<2^31,0<m1,m2<=10。
输出每组数据输出占一行。
样例输入
12 2 3
样例输出
7
来源
爱生活
解析:如果这道题你想用暴力的话就肯定TLE了,这道题用到离散数学中的简单的容斥原理,其实就是集合之间的交并关系,比如A并B=A+B-A交B,算是用到这个公式吧,如果还想更多了解的可以继续百度,知道了这个,就是求m1和m2的最小公倍数,不要想当然的认为m1×m2就是最小公倍数,这个可以利用欧几里德算法来求,剩下就很水了,另外HDU上面也有类似这道题,但是比这个要麻烦,算是这个加强版的,有兴趣的话可以去搜一下哈!
#include <iostream>
using std::endl;
using std::cout;
using std::cin;
//欧几里得求最大公约数
int gcd(int a, int b)
{
if (a % b == 0) {
return b;
} else {
return gcd(b, a % b);
}
}
int main()
{
int n,m1, m2;
while (cin >> n >> m1 >> m2)
{
if (m1 != m2)
cout << (n-1)/m1+(n-1)/m2-(n-1)/((m1*m2)/gcd(m1,m2))<< endl;
else
cout << (n-1)/m1 << endl;
}
return 0;
}
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