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  • hdu1787 GCD Again poj 2478 Farey Sequence 欧拉函数

    hdu1787,直接求欧拉函数

    #include <iostream>
    #include <cstdio>
    using namespace std;
    int n;
    int phi(int n){
    	int ans=n;
    	for(int i=2; i*i<=n; i++)
    		if(n%i==0){
    			ans -= ans / i;
    			while(n%i==0)	n /= i;
    		}
    	if(n>1)	ans -= ans / n;
    	return ans;
    }
    int main(){
    	while(scanf("%d", &n)!=EOF){
    		if(!n)	break;
    		printf("%d
    ",n-phi(n)-1);
    	}
    	return 0;
    }
    

    poj2478,欧拉函数递推,证明可以看这里或者是算法竞赛进阶指南

    (n log n)

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    int n;
    long long phi[1000005];
    void shai(){
    	for(int i=2; i<=1000000; i++)
    		phi[i] = i;
    	for(int i=2; i<=1000000; i++)
    		if(phi[i]==i)
    			for(int j=i; j<=1000000; j+=i)
    				phi[j] = phi[j] / i * (i - 1);
    }
    int main(){
    	shai();
    	for(int i=2; i<=1000000; i++)
    		phi[i] += phi[i-1];
    	while(scanf("%d", &n)!=EOF){
    		if(!n)	break;
    		printf("%lld
    ", phi[n]);
    	}
    	return 0;
    }
    

    线性筛:

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    typedef long long ll;
    int n, pri[1000005], cnt;
    ll phi[1000005];
    bool isp[1000005];
    void shai(){
    	memset(isp, true, sizeof(isp));
    	isp[0] = isp[1] = false;
    	for(int i=2; i<=1000000; i++){
    		if(isp[i])	pri[++cnt] = i, phi[i] = i - 1;
    		for(int j=1; j<=cnt; j++){
    			if(i*pri[j]>1000000)	break;
    			isp[i*pri[j]] = false;
    			if(i%pri[j]==0){
    				phi[i*pri[j]] = phi[i] * pri[j];//感性理解:12:1 5 7 11
    				break;
    			}
    			else	phi[i*pri[j]] = phi[i] * (pri[j] - 1);//积性函数
    		}
    	}
    }
    int main(){
    	shai();
    	for(int i=2; i<=1000000; i++)
    		phi[i] += phi[i-1];
    	while(scanf("%d", &n)!=EOF){
    		if(!n)	break;
    		printf("%lld
    ", phi[n]);
    	}
    	return 0;
    }
    

    上面那种得到了素数,下面这种得到了最小质因子

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    int n, pri[1000005], cnt, val[1000005];
    long long phi[1000005];
    void shai(){
    	for(int i=2; i<=1000000; i++){
    		if(!val[i]){
    			val[i] = i;
    			pri[++cnt] = i;
    			phi[i] = i - 1;
    		}
    		for(int j=1; j<=cnt; j++){
    			if(pri[j]>val[i] || pri[j]>1000000/i)	break;
    			val[i*pri[j]] = pri[j];
    			phi[i*pri[j]] = phi[i]*(i%pri[j]?(pri[j]-1):pri[j]);
    		}
    	}
    }
    int main(){
    	shai();
    	for(int i=2; i<=1000000; i++)
    		phi[i] += phi[i-1];
    	while(scanf("%d", &n)!=EOF){
    		if(!n)	break;
    		printf("%lld
    ", phi[n]);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8067297.html
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