HDU 1796 How many integers can you find(容斥原理)
How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6434 Accepted Submission(s): 1849
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
这道题目是给定了因子,而且还不是互质的,
所以在容斥原理上,就不能简单想乘,要求最小公倍数LCM
另外有可能因子为0
#include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <stdlib.h> #include <math.h> using namespace std; typedef long long int LL; LL gcd(LL a,LL b) { return b?gcd(b,a%b):a; } LL ans,a[15],n,m; void dfs(int id,int flag,int lcm) { lcm=a[id]/gcd(a[id],lcm)*lcm; if(flag) ans+=n/lcm; else ans-=n/lcm; for(int i=id+1;i<m;i++) dfs(i,!flag,lcm); } int main() { while(cin>>n>>m) { n--; for(int i=0;i<m;i++) { scanf("%d",&a[i]); if(!a[i]) i--,m--; } ans=0; for(int i=0;i<m;i++) dfs(i,1,a[i]); printf("%lld\n",ans); } return 0; }