Co-prime HDU - 4135_容斥计数

Code:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=100000+233;
typedef long long ll;
int v[maxn],vis[maxn];
int m[maxn];
int num;          //质因子个数
ll ans=0;
ll A,B;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
void init(ll N){
    int range=(int)sqrt(N),cnt=0;
    for(int i=2;i<=range;++i){
        if(!vis[i])v[++cnt]=i;
        for(int j=1;j<=cnt&&v[j]*i<=range;++i){
            vis[i*v[j]]=1;
            if(i%v[j]==0)break;
        }
    }
    ll t=N;
    for(int i=1;i<=cnt;++i)if(N%v[i]==0)m[++num]=v[i];
    for(int i=1;i<=num&&t!=1;++i)while(t%m[i]==0)t/=m[i];
    if(t>range)m[++num]=t;
}
void dfs(int cur,ll lcm,int id)
{
    if(cur>num)return;
    lcm=m[cur]/gcd(m[cur],lcm)*lcm;
    if(id)
        ans+=(B/lcm)-((A-1)/lcm);
    else ans-=(B/lcm)-((A-1)/lcm);

    for(int i=cur+1;i<=num;++i)
        dfs(i,lcm,!id);
}
int main()
{

    int T;scanf("%d",&T);
    for(int cas=1;cas<=T;++cas)
    {
        memset(v,0,sizeof(v));
        memset(vis,0,sizeof(vis));
        memset(m,0,sizeof(m));
        num=ans=0;
        ll N;
        scanf("%lld%lld%lld",&A,&B,&N);
        init(N);
        for(int i=1;i<=num;++i)
            dfs(i,m[i],1);
        printf("Case #%d: %lld\n",cas,B-A-ans+1);
    }
    return 0;
}