HDU-4746 Mophues




图片说明

可以用埃式筛法处理.

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=5e5+7;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
ll mul(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(ans,a,c);return ans;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
int prime[N],tot=0,p[N],mu[N];
int f[N][20];
ll dp[N][20]={0};
bool vis[N]={0};
void pr(){
    mu[1]=1;p[1]=0;vis[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1,p[i]=1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            p[i*prime[j]]=p[i]+1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<N;i++){
        for(int j=i;j<N;j+=i)
        dp[j][p[i]]+=mu[j/i];
    }
    for(int j=1;j<N;j++){
        for(int i=1;i<20;i++){
            dp[j][i]+=dp[j][i-1];
        }
    }
    for(int j=1;j<N;j++){
        for(int i=0;i<20;i++){
            dp[j][i]+=dp[j-1][i];
        }
    }
}
int main(){
    pr(); 
    int n,m,k,t;cin>>t;
    while(t--){
        sc("%d%d%d",&n,&m,&k);
        if(k>=20){
            printf("%lld\n",1LL*n*m);
            continue;
        }
        if(n>m)swap(n,m);
        ll ans=0;
        for(int i=1,last;i<=n;i=last+1){
            last=min(n/(n/i),m/(m/i));
            ans+=(dp[last][k]-dp[i-1][k])*(n/i)*(m/i);
        }
        printf("%lld\n",ans);
    }
}