How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6429 Accepted Submission(s): 1847
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
思路：容斥原理；需要注意的是给你的数有可能包含0，只要把0换成比n-1大的数或者去掉就行；
还有求的是<n的，那么这时麻烦的地方就是要判断整除，所以转变下就是求(<=n-1)就行这时不需要判断是否整除。
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<stdlib.h> 5 #include<string.h> 6 #include<vector> 7 #include<queue> 8 #include<stack> 9 using namespace std; 10 long long gcd(long long n,long long m); 11 int ans[20]; 12 int main(void) 13 { 14 int i,j,k; 15 int n,m; 16 while(scanf("%d %d",&n,&m)!=EOF) 17 { 18 int sum=0; 19 n=n-1; 20 for(i=0; i<m; i++) 21 { 22 scanf("%d",&ans[i]); 23 } 24 for(i=0; i<m; i++) 25 { 26 if(ans[i]==0) 27 ans[i]=n+1; 28 } 29 for(i=1; i<=(1<<m)-1; i++) 30 { 31 int cnt=0; 32 long long an=1; 33 int flag=0; 34 for(j=0; j<m; j++) 35 { 36 if(i&(1<<j)) 37 { 38 cnt++; 39 long long cc=gcd(an,(long long)ans[j]); 40 an=an/cc*ans[j]; 41 if(an>n) 42 { 43 flag=1; 44 break; 45 } 46 } 47 } 48 if(flag) 49 continue; 50 else 51 { 52 if(cnt%2) 53 sum+=n/(int)an; 54 else sum-=n/(int)an; 55 } 56 } 57 printf("%d ",sum); 58 } 59 return 0; 60 } 61 long long gcd(long long n,long long m) 62 { 63 if(m==0) 64 return n; 65 else if(n%m==0) 66 return m; 67 else return gcd(m,n%m); 68 }