HDU 4135 容斥原理

思路：

直接容斥

//By SiriusRen
#include <cstdio>
using namespace std;
#define int long long
const int N=;
int cases,a,b,n,tp,s[N];
int solve(int r){
    int ans=;
    for(int i=;i<(<<tp);i++){
        int tmp=,t=;
        for(int j=;j<tp;j++)
            if(i&(<<j))tmp*=s[j],t++;
        t=t&?:-;
        ans+=r/tmp*t;
    }
    return r-ans;
}
signed main(){
    scanf("%lld",&cases);
    for(int I=;I<=cases;I++){
        tp=;
        scanf("%lld%lld%lld",&a,&b,&n);
        for(int i=;i*i<=n;i++){
            if(n%i)continue;
            while(n%i==)n/=i;
            s[tp++]=i;
        }if(n!=)s[tp++]=n;
        printf("Case #%lld: %lld\n",I,solve(b)-solve(a-));
    }
}