[莫比乌斯反演+数位DP]计蒜客 阿里云秘钥池

$$f[i][j]=\sum_{k=1}^{P-1}f[i-1][k]*e(gcd(j,k))$$
$$f[i][j]=\sum_{k=1}^{P-1}f[i-1][k]\sum_{d|j,d|k}\mu(d)$$
$$f[i][j]=\sum_{d|j}\mu(d)*\sum_{k=1}^{\lfloor\frac{p-1}{d}\rfloor}f[i-1][kd]$$

然后就数位DP一下

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100013
int T,P,mu[N],prim[N],c[70];
char p[N];
ll l,r,sum,f[70][N];
void pre(void){
    ll t;mu[1]=1;
    for(int i=2;i<=N;++i){
        if(!p[i])prim[++prim[0]]=i,mu[i]=-1;
        for(int j=1;j<=prim[0]&&(t=1ll*i*prim[j])<=100000;++j){
            p[t]=1;
            if(i%prim[j]==0){
                mu[t]=0;break;
            }
            mu[t]=-mu[i];
        }
    }
    return ;
}
int gcd(int a,int b){
    return a%b==0?b:gcd(b,a%b);
}
ll calc(ll x){
    ll sum=0,a=x;
    int tot=0;
    while(a)c[++tot]=a%P,a/=P;  
    for(int i=tot;i;--i){
        for(int j=1;j<c[i];++j)
            if(i==tot||(gcd(j,c[i+1])==1))
                sum+=f[i][j];
        if(c[i]==0)break;
        if(i!=tot&&gcd(c[i],c[i+1])!=1)break;
    }
    for(int i=tot-1;i;--i)
        for(int j=1;j<P;++j)
            sum+=f[i][j];
    return sum;
}
int main(void){
    register int i,j,k,len;
    ll x;
    pre(),scanf("%d",&T);
    while(T--){
        scanf("%lld%lld%d",&l,&r,&P);
        x=r+1,len=0;while(x)x/=P,++len;
        for(i=1;i<P;++i)f[1][i]=1;
        for(i=2;i<=len;++i){
            for(j=1;j<P;++j)f[i][j]=0;
            for(j=1;j<P;++j){
                sum=0;
                for(k=j;k<P;k+=j)
                    sum+=f[i-1][k];
                for(k=j;k<P;k+=j)
                    f[i][k]+=mu[j]*sum;
            }
        }
        printf("%lld\n",calc(r+1)-calc(l));
    }
    return 0;
}