HDOJ 1796 枚举

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1219Accepted Submission(s): 282

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

具体做法是去枚举集合，总共会有1<<m种，这里运用到了二进制位运算。

比起容斥原理中的DFS，我更喜欢这种计算方法。

#include<stdio.h> __int64 a[15]; __int64 gcd(__int64 a,__int64 b) { if(b==0) return a; else return gcd(b,a%b); } __int64 lcm(__int64 a,__int64 b) { __int64 g=gcd(a,b); return a*b/g; } int main() { __int64 n,m,i,j,k,s,ans,num; while(scanf("%I64d%I64d",&n,&m)!=EOF) { num=0; for(i=0;i<m;i++) { scanf("%I64d",&a[i]); if(a[i]>0) a[num++]=a[i]; } ans=0,m=num; for(i=1;i<(1<<m);i++) { s=1; k=0; for(j=0;j<m;j++) { if((i>>j)&1) { s=lcm(s,a[j]); k++; } } if(k&1) ans=ans+(n-1)/s; else ans=ans-(n-1)/s; } printf("%I64d\n",ans); } return 0; }