1 //phi(a)=a*(a1-1)*(a2-1)*(a3-1)*...*(an-1)/(a1*a2*a3*...*an); 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 typedef __int64 LL; 8 LL phi(LL a){//phi 9 LL temp=a; 10 for(LL i=2;i*i<=a;i++) 11 if(a%i==0){ 12 while(a%i==0) a/=i; 13 temp=temp/i*(i-1); 14 // cout<<"a="<<a<<" i="<<i<<endl; 15 } 16 //cout<<"a="<<a<<endl; 17 if(a!=1) temp=temp/a*(a-1); 18 return temp; 19 } 20 int main(){ 21 LL a; 22 while(cin>>a) 23 cout<<"phi(a)="<<phi(a)<<endl; 24 return 0; 25 }
快速乘方 求a的k次方mod b
1 #include<iostream> 2 #include<cstdlib> 3 #include<cstring> 4 using namespace std; 5 typedef __int64 LL; 6 LL modd(LL a,LL k,LL b){ 7 int temp,ans; 8 ans=1; 9 temp=a; 10 while(k!=0){ 11 if(k%2) ans=ans*temp%b; 12 temp=temp*temp%b; 13 k/=2; 14 } 15 return ans; 16 } 17 int main(){ 18 LL a,b,k; 19 while(cin>>a>>b>>k) 20 cout<<modd(a,k,b)<<endl; 21 return 0; 22 }
原文:http://www.cnblogs.com/acplayfacm/p/3864654.html