ahu-557容斥原理

用容斥求1-n中能被2-i中素数整除的个数ans其中 i*i<=n；然后结果为n-ans-1;

#include<cstdio>
int pri[10005],pri2[10005],i1,n,ans;
void toGetPrim(){
    for(int i=2;i<=10001;i++)
     if(!pri[i]){
        for(int j=2;j*i<=10001;j++)
         pri[i*j] = 1;
       }
     i1 = 0;
     for(int i=2;i<=10001;i++)
     if(!pri[i]){
       pri2[i1] = i;
       i1++;
      }
}
void dfs(int re,int all,int pre){
      if(re!=1)
      {
      if(all==1)ans-=1;
      if(all%2)ans+=n/re;
      else ans-=n/re;
      }
      for(int i=pre;i<i1;i++){
        long long g = (long long)re*pri2[i];
        if(g>n)return;
        dfs(g,all+1,i+1);
      }
}
int main()
{
    toGetPrim();
    while(scanf("%d",&n)&&n)
    {
        ans = 0;
        dfs(1,0,0);
        printf("%d\n",n-1-ans);
    }
    return 0;
}