NOIP Simulation---1. Angry ljj (Anger)

LJJ just finished a class! This class is a math class! He knows that. The addition and subtraction is a first-class operation, multiplication belongs to a two-level operation, the power is a three-level operation, and the priority of power > Multiplication Priority > The priority of the addition and subtraction (this is the math class in several grades). However, from the previous set of papers + Previous question, we know that LJJ is a person who is always a whim and not enough IQ (that is, he came up with a question for you). He invented a four-level operation, and we're going to use the sign # to indicate (no other symbol is found). We know A*b=a+a+a+...+a (plus b), a^b=a*a*a*a*...*a (multiply B), then a#b=a^a^a^a^...^a (Power operation B), natural, #的优先级比幂的优先级高. So, LJJ asked you to help him find a#b mod 1000000007. (PS: This is the most simple question of this paper)

Title Description

See topic Background

Input/output format

Input format:

Enter only 1 lines, that is, a, B.

Output format:

Output is only 1 lines, i.e. a#b mod 1000000007.

Input and Output Sample input example # #:

3 5

Sample # # of output:

968803245

Description

First of all, the sample answer is not mod is actually 4.4342648824303776994824963061915e+38 (from the person who malicious)

Then, the data range:

For 20% of data, a<=1000,b<=1000

For 50% of data, a<=10^16,b<=10000

For 100% of data, a<=10^16,b<=10^16

Analysis: http://blog.163.com/eden_284914869/blog/static/2522460782016799444725/

Good number theory nature (to be proved): a^ (b mod (prime-1)) mod prime = a^ B mod Prime (Prime is a prime number)

A^a^a......^a (b-time a) mod prime = a^ (a^ (b-1)) mod prime=a^ (a^ (b-1) mod (prime-1)) mod prime

Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<climits>#include<cstring>#include<string>#include<Set>#include<map>#include<queue>#include<stack>#include<vector>#include<list>#include<bitset>#defineRep (I,m,n) for (i=m;i<=n;i++)#defineRSP (It,s) for (Set<int>::iterator It=s.begin (); It!=s.end (); it++)#defineVI vector<int>#definePII pair<int,int>#defineMoD 1000000007#defineINF 0x3f3f3f3f#definePB Push_back#defineMP Make_pair#defineFi first#defineSe Second#definell Long Long#definePi ACOs (-1.0)Const intmaxn=1e5+Ten;Const intdis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};using namespacestd;ll gcd (ll p,ll q) {returnq==0? P:GCD (q,p%q);} ll Qpow (ll p,ll q,ll mo) {ll F=1; while(q) {if(q&1) f=f*p%mo;p=p*p%mo;q>>=1;}returnF;}intN,m,k,t;ll A, b;intMain () {inti,j; scanf ("%lld%lld",&a,&b); printf ("%lld\n", Qpow (A%mod,qpow (a% (mod-1), B-1, mod-1) , MoD); //System ("pause");    return 0;}

NOIP Simulation---1. Angry ljj (Anger)