HDU 1796 How many integers can you find（组合数学-容斥原理）

How many integers can you find

Problem Description

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10},
all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

 

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

 

Output

  For each case, output the number.

 

Sample Input

12 2
2 3

 

Sample Output

7

 

Author

wangye

 

Source

2008
“Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

 

Recommend

wangye

题目大意：
输入N，和M个数，求出1～N-1中，有几个数能被M个数任意一个整除。

解题思路：
利用容斥原理。被任意一个整除的个数 = 同时被一个整除 - 同时被两个整除 + 同时被三个整除 - ...
A∪B∪C = A+B+C - A∩B - B∩C - C∩A +A∩B∩C

区间[1,N]中能同时被x1, x2, ..., xm整除的数的个数 = N / lcm(x1, x2, ..., xm)

可以用位运算，简化枚举状态的过程。

参考代码：

#include <iostream>
#include <vector>
using namespace std;

vector<int> a;
int n, m;
long long ans;

long long gcd(long long x, long long y) {
return y > 0 ? gcd(y, x % y) : x;
}

long long lcm(long long x, long long y) {
return x * y / gcd(x, y);
}

void init() {
a.clear();
ans = 0;
}

void input() {
for (int i = 0; i < m; i++) {
int x;
cin >> x;
if (x > 0) {
a.push_back(x);
}
}
m = a.size();
}

void work() {
for (int i = 1; i < (1 << m); i++) {
long long x = 1;
int cnt = 0;
for (int j = 0; j < m; j++) {
if (i & (1 << j)) {
cnt++;
x = lcm(x, a[j]);
}
}
if (cnt & 1) {
ans += (n - 1) / x;
} else {
ans -= (n - 1) / x;
}
}
}

void output() {
cout << ans << endl;
}

int main() {
ios::sync_with_stdio(false);
while (cin >> n >> m) {
init();
input();
work();
output();
}
return 0;
}