FZU 1759 Super A^B mod C 指数循环节

Problem 1759 Super A^B mod C

Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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
思路：指数循环节，数据水，并没有B<C&&A,C不互质的；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e5+10,M=1e6+10,inf=1e9+10;
const int mod=1000000007;
#define MAXN 10000010
ll quickpow(ll x,ll y,ll z)
{
ll ans=1;
while(y)
{
if(y&1)
ans*=x,ans%=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;
}
char a[M];
int main()
{
ll x,y,z,i,t;
while(~scanf("%I64d%s%I64d",&x,a,&z))
{
t=strlen(a);
ll p=phi(z);
ll ans=0;
for(i=0;i<t;i++)
ans=(ans*10+a[i]-'0')%p;
ans+=p;
printf("%I64d\n",quickpow(x,ans,z));
}
return 0;
}