Submission #6925636

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Source Code Expand

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <tuple>
#include <stdio.h>
#include <bitset>
#include <limits.h>
#include <complex>
#include <deque>
#include <iomanip>
#include <list>
#include <time.h>
using namespace std;
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
#define en cout << endl //セミコロンつけろ
//#define MM 1000000000
//#define MOD MM+7
const int MM = 1e9;
const int MOD = MM+7;
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
// 'A' = 65, 'Z' = 90, 'a' = 97, 'z' = 122
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
template<typename T>
T LCM(T x,T y){
    T gc = GCD(x,y);
    return x*y/gc;
}

long long Pow(long long base,long long x,long long mod){
    //マクロ定義のままMODを使ったらエラーを起こした(多分long long の定義の問題(const long long などに要変更))
    if (x == 0) return 1;
    //指数が奇数の時、x^(2n+1) -> x * x^(2n)と変換
    if (x%2 == 1) return base * Pow(base,x-1,mod) % mod;
    //指数が偶数なら x^(2n) -> x^(2n) * x^(2n)と変換
    long long y = Pow(base, x/2,mod);
    return y * y % mod;
}

int main(){
    long long N,M,P; cin >> N >> M >> P;
    cout << Pow(N,P,M) << endl;
}

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