思路:
因为指数非常大,用long long 都存不下, 所以要用数组存, 然后在通过欧拉降幂取模是指数变小,在通过快速幂求结果
注:这里只用到了欧拉降幂中 指数大于等于模数的情况
#include<iostream>
#include<cstring>
#define ll long long
using namespace std;
const int maxn = 1000005;
ll qpow(ll a, ll n, ll mod)
{
ll ret = 1;
while(n)
{
if(n%2) ret = (ret%mod)*(a%mod)%mod;
a = (a%mod)*(a%mod)%mod;
n/=2;
}
return ret;
}
ll phi(ll x)
{
ll ret = x;
for(ll i = 2;i*i <= x;i++)
{
if(x%i == 0)
{
ret = ret*(i-1)/i;
while(x%i == 0)x/=i;
}
}
if(x > 1)ret = ret*(x-1)/x;
return ret;
}
int main()
{
ios_base::sync_with_stdio(false);
ll a, c;
char b[maxn];
while(cin>>a>>b>>c)
{
ll n = 0;
ll len = strlen(b);
if(len<= 10)
{
for(int i = 0;i < len;i++)
{
n = n*10+b[i]-'0';
}
}
else // 指数长度大于10 肯定大于模数
{
ll ph = phi(c);
for(int i = 0;i < len;i++)
{
n = (n*10+b[i] - '0')%ph;
}
n += ph;
}
cout<<qpow(a, n, c)<<endl;
}
return 0;
}