codeforces 1007B Pave the Parallelepiped

https://blog.csdn.net/qq_34454069/article/details/81043195 

【组合计数】

 

 

 

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define SF scanf
#define PF printf
#define MAXN 100010
using namespace std;
int n;
typedef long long ll;
ll f[MAXN],num[MAXN];
int primes[MAXN],isprime[MAXN],tot;
void prepare(){
    f[1]=1;
    for(int i=2;i<=100000;i++){
        if(isprime[i]==0){
            primes[++tot]=i;
            f[i]=2;
            num[i]=1;
        }
        for(int j=1;j<=tot&&i*primes[j]<=100000;j++){
            isprime[i*primes[j]]=1;
            if(i%primes[j]==0){
                num[i*primes[j]]=num[i]+1;
                f[i*primes[j]]=f[i]/(num[i]+1ll)*(num[i*primes[j]]+1ll);
                break;
            }
            f[i*primes[j]]=f[i]*f[primes[j]];
            num[i*primes[j]]=1;
        }
    }
}
int gcd(int x,int y){
    if(y==0)
        return x;
    return gcd(y,x%y);
}
int main(){
    prepare();
    int t,a,b,c;
    SF("%d",&t);
    for(int i=1;i<=t;i++){
        SF("%d%d%d",&a,&b,&c);
        //PF("{%lld %lld %lld}
",f[a],f[b],f[c]);
        //PF("{%d %d %d}",gcd(a,b),gcd(b,c),gcd(a,c));
        int ab=gcd(a,b),bc=gcd(b,c),ac=gcd(a,c);
        ll ans1=f[a]*f[b]*f[c]; 
        ll ans2=(f[ab]*f[ab]-f[ab])/2ll*f[c];
        ll ans3=f[a]*(f[bc]*f[bc]-f[bc])/2ll;
        ll ans4=f[b]*(f[ac]*f[ac]-f[ac])/2ll;
        int td=gcd(gcd(a,c),b);
        ll ans5=(f[td]*f[td]-f[td])+f[td]*(f[td]-1ll)*(f[td]-2ll)/6ll*4ll;
        ll ans6=(f[ab]-f[td])*f[td]*(f[td]-1ll)/2ll;
        ll ans7=(f[bc]-f[td])*f[td]*(f[td]-1ll)/2ll;
        ll ans8=(f[ac]-f[td])*f[td]*(f[td]-1ll)/2ll;
        ll ans9=(f[ab]-f[td])*(f[bc]-f[td])*(f[ac]-f[td]);
        ll ans=ans1-ans2-ans3-ans4+ans5+ans6+ans7+ans8-ans9;
        /*if(ans==212)
            PF("[%d,%d,%d]
",a,b,c);*/
        //PF("%lld %lld %lld %lld %lld",ans1,ans2,ans3,ans4,ans5); 
        PF("%I64d
",ans);
    }
}