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 }
