POJ2480 Longge's problem 

  1 /*
  2 POJ2480 Longge's problem
  3 http://poj.org/problem?id=2480
  4 数论 欧拉函数 线性筛 
  5 
  6 求sigma_i=1^n{gcd(i,n)}
  7 <=>sigma_d|n{d*phi(n/d)}
  8  *
  9  *
 10  *
 11  *
 12  */
 13 #include <cstdio>
 14 #include <algorithm>
 15 #include <cmath>
 16 #define new
 17 #ifdef OLD
 18 //250ms
 19 using namespace std;
 20 const long long Nmax=1000005LL;
 21 long long n;
 22 int is_prime[Nmax];
 23 int prime[Nmax];
 24 int cnt;
 25 long long phi[Nmax];
 26 long long oula(long long n)
 27 {
 28     if(n==1LL)
 29         return 1LL;
 30     if(n<Nmax)
 31         return phi[n];
 32     long long ret=1LL,i;
 33     for(i=2LL;i*i<=n;i++)
 34     {
 35         if(n%i==0LL)
 36         {
 37             n/=i,ret*=i-1LL;
 38             while(n%i==0LL) n/=i,ret*=i;
 39         }
 40     }
 41     if(n>1LL) ret*=n-1LL;
 42     return ret;
 43 }
 44 
 45 void get()
 46 {
 47     for(int i=2;i<Nmax;i++)
 48         is_prime[i]=1;
 49     for(long long i=2;i<Nmax;i++)
 50     {
 51         if(is_prime[i])
 52         {
 53             prime[++cnt]=i;
 54             phi[i]=(long long)(i-1);
 55         }
 56         for(int j=1;j<=cnt;j++)
 57         {
 58             if(i*prime[j]>=Nmax)
 59                 break;
 60             is_prime[i*prime[j]]=0;
 61             if(i%prime[j]==0)
 62             {
 63                 phi[i*prime[j]]=phi[i]*(long long)prime[j];
 64                 break;
 65             }
 66             else
 67                 phi[i*prime[j]]=phi[i]*(long long)(prime[j]-1);
 68         }
 69     }
 70 }
 71 
 72 int main()
 73 {
 74     // freopen("1.in","r",stdin);
 75     //scanf("%lld%lld",&a,&b);
 76     //printf("%lld\n",gcd(a,b));
 77     get();
 78     while(scanf("%lld",&n)==1)
 79     {
 80         //n=read();
 81         //printf("%lld\n",n);
 82         //if(n==0LL)
 83             //break;
 84         if(n<Nmax && is_prime[n])
 85         {
 86             printf("%lld\n",2LL*n-1LL);
 87             continue;
 88         }
 89         long long ans=0LL;
 90         for(long long i=1LL;i*i<=n;i++)
 91         {
 92             if(n%i==0LL)
 93                 ans+=i*oula(n/i)+n/i*oula(i);
 94             if(i*i==n)
 95                 ans-=i*oula(n/i);
 96         }
 97         printf("%lld\n",ans);
 98     }
 99     
100     //for(long long  n=1;n<=100;n++)
101     //{
102         //long long ans=0LL;
103         //for(long long i=1LL;i<=n;i++)
104             //ans+=gcd(i,n);
105         //printf("%d:%d\n",n,ans);
106         //long long m=n;
107         ////for(long long i=1;prime[i])
108     //}
109     return 0;
110 }
111 
112 #endif
113 
114 #ifdef new
115 //32ms
116 
117 using namespace std;
118 int main()
119 {
120     long long n;
121     while (scanf("%lld", &n) != EOF) 
122     {
123         long long i, sqr, p, a, ans;
124         ans = n;
125         sqr = floor(sqrt(n*1.0));
126         for (i = 2; i <= sqr; i++) 
127         {
128             if (n%i == 0) 
129             {
130                 a = 0;
131                 p = i;
132                 while (n%p == 0) 
133                 {
134                     a++;
135                     n /= p;
136                 }
137                 sqr = floor(sqrt(n*1.0));
138                 ans = ans + ans*a*(p-1)/p;
139             }
140         }
141         if (n!=1) 
142             ans = ans + ans*(n-1)/n;
143         printf("%lld\n", ans);
144     }
145     return 0;
146 }
147 
148 #endif

 

posted @ 2017-03-15 00:36  BBBob  阅读(211)  评论(0)    收藏  举报