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
处理大数的方法挺经典的，把别人代码贴出来以示膜拜
![]()
//A^B %C=A^( B%phi(C)+phi(C) ) %C
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include<string>
#include<cmath>
using namespace std;
typedef __int64 ll;
int phi(
int x)
{
int i,j;
int num = x;
for(i = 2; i*i <= x; i++)
{
if(x % i == 0)
{
num = (num/i)*(i-1);
while(x % i == 0)
{
x = x / i;
}
}
}
if(x != 1) num = (num/x)*(x-1);
return num;
}
ll quickpow(ll m,ll n,ll k)
{
ll ans=1;
while(n)
{
if(n&1) ans=(ans*m)%k;
n=(n>>1);
m=(m*m)%k;
}
return ans;
}
char tb[1000015];
int main()
{
ll a,nb;
int c;
while(
scanf("
%I64d%s%d",&a,tb,&c)!=EOF)
{
int PHI=phi(c);
ll res=0;
for(
int i=0;tb[i];i++)
{
res=(res*10+tb[i]-'0');
if(res>c)
break;
}
if(res<=PHI)
{
printf("
%I64d\n",quickpow(a,res,c));
}
else
{
res=0;
for(
int i=0;tb[i];i++)
{
res=(res*10+tb[i]-'0')%PHI;
}
printf("
%I64d\n",quickpow(a,res+PHI,c));
}
}
return 0;
}