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how i solve this problem (A^(B^C))mod 1e9+7
By ypizarroza, 9 years ago, In English

please someone that help me whith this problem http://www.spoj.com/problems/POWERUP/ i'm using the fermat little theorem a^p-1=a (mod p). Grace in advance

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9 years ago, # |
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By Fermat theorem a^phi(m) == a (mod m) => a^(phi(m) — 1) == 1 (mod m). In this problem m == 10^9 + 7 is prime, so phi(m) == m — 1.

So A^(B^C) (mod m) == A^(B^C mod (phi(m) — 1)) (mod m) == A^(B^C mod (m — 2)) (mod m).

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    9 years ago, # ^ |
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    Yes, I'm applying that theory,but the judge sentence is wrong answer. I would appreciate if you provide me extreme cases in which my code it can fail

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    9 years ago, # ^ |
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    You are wrong, aφ(m) = 1.

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    9 years ago, # ^ |
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    It's Euler theorem, not Fermat's theorem...

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9 years ago, # |
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There are some tricky cases when a mod p == 0

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    9 years ago, # ^ |
    ← Rev. 2   Vote: I like it -8 Vote: I do not like it 
    I have proven to these cases. Please one user that has AC this problem. please  look my code:
#include <bits/stdc++.h>
#define ifc(x)(flag[x>>6]&(1<<((x>>1)&31)))
#define isc(x)(flag[x>>6]|=(1<<((x>>1)&31)))
using namespace std;
typedef long long i64;
#define MOD 1000000007
typedef long long i64;
i64 multiply(i64 a,i64 b, i64 mod){
 i64 rx=0,sx=0;
  for (int bx=0; b>>bx>0; ++bx){
    sx+=(bx)?sx:a;
    if(sx>=mod) sx-=mod;
    rx+=((b>>bx)&1)?sx:0;
    while(rx>=mod) rx-=mod;
  }
  return rx;
}
i64 modpow(i64 a,i64 b, i64 mod){
  i64 rx=1,sx=0;
  for(int bx=0; b>>bx>0; ++bx){
    sx=(bx)?multiply(sx,sx,mod):a;
    rx=((b>>bx)&1)?multiply(rx,sx,mod):rx;
  }
  return rx;
}
i64 sol;
int cas,a,b,c;
int main(){

	      while(1){
	      scanf("%d %d %d",&a,&b,&c);
	      if(a+b+c==-3)break;
		  i64 x=modpow(b,c,MOD-1);
		  printf("%lld\n",modpow(a,x,MOD));
	      }
}

    ``

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