hdu1796(二进制枚举)
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
典型的二进制枚举的题,数据开的不大,所以容易二进制枚举很容易做到!这题有点坑的地方在于the M integer are non-negative and won’t exceed 20.是说的非负数,也就是说可能会有0出现,这个条件没看到刚开始死活不a啊,思考数据方面的能力真的很需要加强啊。
其次这个题还是典型的容斥原理!本是水题一道,唉。。。。
#include <iostream> #include <stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; } ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } int main() { int n,m;ll a[30]; ll b; while(~scanf("%d%d",&n,&m)) { n--;int k=m+1; for(int i=0;i<m;i++) { scanf("%lld",&a[i]);if(a[i]==0)k=i; } if(k!=m+1) { m--;swap(a[k],a[m]); } ll s=0; for(ll i=1;i<1<<m;i++) { b=1; ll t=i,num=0,nn=0,flag=0; while(1) { int x=t%2; if(x) { b=lcm(b,a[num]);nn++; } t/=2;if(t==0)break; num++; } if(nn&1) s+=n/b; else s-=n/b; }printf("%d\n",s); } return 0; }
持续更新博客地址:
blog.csdn.net/martinue