hdu1796 How many integers can you find (容斥原理)

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 22 3

 

Sample Output

7

 

Author

wangye

题意：

问有多少个小于n的数能被集合里的数整除。集合拥有m个元素。

自己写的DFS容斥， 写得很粗糙， 在求最大公因数的时候没有用long long ，爆了int， 无情wa，于是破罐子破摔把所有的int都改成 long long就过了。后面附上大神代码， 十分精简

#include<iostream>#include<cstdio>#include<algorithm>#define ll long longusing namespace std;int n, m, num[15];long long  sum, gg;int cmp(int k1, int k2)  {   int a = max(k1, k2);    int b = min(k1, k2);        if(b == 0) return a;int temp;    while(b != 0)      {          temp = a % b;          a = b;          b = temp;      }          return (1LL * k1 * k2) / a ;  }  ll ans(int x){if(x == 0) return 0;ll te = n / x, mod = n % x;if(mod == 0)  return te - 1; return te;} void DFS(int po, int te, int n, int co){      int cc = te;if(n == co){  //cout<<ans(te)<<endl;      gg +=ans(te);     // cout<<"te="<<te<<endl;      //cout<<gg<<endl;    }     for(int i = 1; (po + i) < m; i++){        te = cmp(cc,num[po + i]);DFS(po + i, te, n + 1, co);}   } int main(){while(scanf("%d", &n) != EOF){ scanf("%d", &m);for(int i = 0; i < m; i++){scanf("%d", num + i);if(num[i] == 0){  i--;  m--;}}    sum = 0; int f = -1;for(int co = 1; co <= m; co++) { f *= -1; for(int i = 0; i < m; i++){     gg = 0;     DFS(i, num[i], 1, co);            sum+= f * gg;        }}  cout<<sum<<endl;    }return 0;}

附上大神代码

#include<iostream>#include<cstdio>#include<algorithm>#define ll long longusing namespace std;ll ans;int n, m, num[15];int gcd(ll a, ll b){return b == 0 ? a : gcd(b, a % b);} void DFS(int cur, ll lcm, int vis){   lcm = num[cur] / gcd( num[cur], lcm) * lcm;   if(vis % 2)        ans += (n - 1) / lcm;   else         ans -= (n - 1) / lcm;           for(int i = cur + 1; i < m; i++)        DFS(i, lcm, vis + 1);}int main(){while(scanf("%d%d", &n, &m) != EOF){for(int i = 0; i < m; i++){scanf("%d", num + i);if(num[i] == 0)   i-- , m--; }ans = 0;for(int i = 0; i < m ; i++){DFS(i, num[i], 1);}cout<<ans<<endl;}    return 0;}