ZOJ 2836 Number Puzzle

Number Puzzle

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 2836
64-bit integer IO format: %lld      Java class name: Main

Given a list of integers ($A_1, A_2, ..., A_n$), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N ($1 \leq N \leq 10$) and $M (1 \leq M \leq 200000000)$, and the second line contains $A_1, A_2, ..., A_n(1 \leq A_i \leq 10, for\ i = 1, 2, ..., N).$

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2
2 3 7
3 6
2 3 7

Sample Output

1
4

Source

Zhejiang University Local Contest 2007

Author

MAO, Yiqiang

解题：容斥，求是集合中某些数的LCM的倍数且不超过m的数的数量

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 20;
 5 int a[maxn],n,m;
 6 LL gcd(LL a,LL b) {
 7     return b?gcd(b,a%b):a;
 8 }
 9 LL LCM(LL a,LL b) {
10     return a/gcd(a,b)*b;
11 }
12 int main() {
13     while(~scanf("%d%d",&n,&m)) {
14         for(int i = 0; i < n; ++i)
15             scanf("%d",a+i);
16         LL ret = 0;
17         for(int i = 1; i < (1<<n); ++i) {
18             int cnt = 0;
19             LL x = 1;
20             for(int j = 0; j < n; ++j) {
21                 if((i>>j)&1) {
22                     x = LCM(x,a[j]);
23                     cnt++;
24                 }
25             }
26             if(cnt&1) ret += m/x;
27             else ret -= m/x;
28         }
29         printf("%lld\n",ret);
30     }
31     return 0;
32 }