Cod sursa(job #1732968)

Utilizator danutbodbodnariuc danut danutbod Data 23 iulie 2016 13:11:16
Problema Ridicare la putere in timp logaritmic Scor 100
Compilator cpp Status done
Runda Arhiva educationala Marime 1.51 kb
//#include <fstream> //var III cu desc. in baza 2
//const long long  MOD = 1999999973;
//using namespace std;
//ifstream f("lgput.in");
//ofstream g("lgput.out");
//long long  n,p;
//long long putere(long long n, long long p) {
//    long long sol = 1;
//    n %= MOD;
//    for(; p; p >>= 1) {
//        if(p & 1)  sol = (sol*n)%MOD;
//        n= (n*n)%MOD;
//    }
//    return sol;
//}
//int main()
//{
//  f>>n>>p;
//  g<<putere(n,p);
//  return 0;
//}

//#include <fstream> //var II
//using namespace std;
//const int m = 1999999973;
//ifstream fi("lgput.in");
//ofstream fo("lgput.out");
//int i, n, p;
//long long putere(long long  n,long long p){//calc. (a^p)%m
//    long long sol=1;
//    for (i=0;(1<<i)<=p;i++)  // Luam toti biti lui p la rand
//    {
//        if ( ((1<<i) & p) > 0) // Daca bitul i din p este 1 atunci adaugam n^(2^i) la solutie
//            sol= (sol * n) % m;
//            n=(n * n) % m; // Inmultim n cu n ca sa obtinem n^(2^(i+1))
//    }
//  return sol;
//}
//int main()
//{
//    fi>>n>>p;
//    fo<< putere(n,p); // Afisam solutia
//}

#include <fstream> //var I recursiva
using namespace std;
const long long c = 1999999973;
ifstream f("lgput.in");
ofstream g("lgput.out");
long long  n,p;
long long  putere(long long  n,long long  p)
{
    long long r;
    if (p==0) return 1;
    else
       if (p%2==0){r=putere(n,p/2)%c;return(r*r)%c;}
       else  return (putere(n,p-1)*n)%c;
}

int main()
{
  f>>n>>p;
  g<<putere(n,p);
    return 0;
}