欧拉降幂

当求a的b次幂%p时，如果在求b的时候被迫要取余的话，那么就要用到欧拉降幂

 

 就像这个是的就是在就fn的时候不得不取余，因为fn是指数上的如果直接取余的话肯定会wa,所以要用到欧拉降幂，就是在求fn的时候mod(1e9+7-1)

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=1e6+100;
ll f[maxn];
ll a,b,m,n;
ll qpow(ll a,ll b){
    ll ans=1;
    a%=mod;
    while(b){
        if(b&1){
            ans=(ans*a)%mod;
        }
        a=(a*a)%mod;
        b/=2; 
    }
    return ans;
}
void inint(){
    f[1]=1;
    f[2]=1;
    for(int i=3;i<=110000;i++){
        f[i]=(a*f[i-1]+b*f[i-2])%(mod-1); 
    }    
}
int main(){
    cin>>a>>b>>m>>n;
    inint();
    ll ans=qpow(m,f[n])%mod;
    cout<<ans<<endl;
}

Description

 

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4
2 10 1000

Sample Output

1
24

Hint

 

Source

FZU 2009 Summer Training IV--Number Theory

 

题意：A^B mod C

题解：

降幂公式 phi() 为欧拉函数

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll __int64
#define mod 10000000007
using namespace std;
char a[1000006];
ll x,z;
ll quickpow(ll x,ll y,ll z)
{
    ll ans=1;
    while(y)
    {
        if(y&1)
            ans=ans*x%z;
        x=x*x%z;
        y>>=1;
    }
    return ans;
}
ll phi(ll n)
{
    ll i,rea=n;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            rea=rea-rea/i;
            while(n%i==0)
                n/=i;
         }
    }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}
int main()
{
    while(scanf("%I64d %s %I64d",&x,a,&z)!=EOF)
    {
        ll len=strlen(a);
        ll p=phi(z);
        ll ans=0;
        for(ll i=0;i<len;i++)
            ans=(ans*10+a[i]-'0')%p;
        ans+=p;
        printf("%I64d
",quickpow(x,ans,z));
    }
    return 0;
}