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Андрей
Asked:2022-02-01 10:34:16 +0000 UTC2022-02-01 10:34:16 +0000 UTC

模幂未通过所有测试

  • 772

https://www.eolymp.com/en/problems/5336

#include <stdio.h>
#include <iostream>
 
using namespace std;
 
int main()
{
    int a, n, rezult = 1, i;
    int chislo;
    cin >> a;
    cin >> n;
    chislo = 0;
    for (i = 0; i < n; i++)
    {
        rezult *= a;
    }
    chislo = rezult % 1000000007;
    cout << chislo;
    return 0;
}
c++

1 个回答

  1. Best Answer

    二进制取幂和“非常长的算术”:) 从长数中取模。

    #include <iostream> 
#include <string>
using namespace std; 

const long long mod = 1000000007;
 
long long int  powLL(long long int  x, long long int  n) 
{ 
   long long int  result = 1; 
    x %= mod;
    if(x == 0) return 0;
    while (n)
    { 
        if (n & 1)  result = (result * x) % mod; 
        n /= 2; 
        x = (x * x) % mod; 
    } 
    return result; 
} 
 
long long int  powStrings(string sa, string sb) 
{
    long long int  a = 0, b = 0;  
    for (int i = 0; i < sa.size(); i++) a = (a * 10 + (sa[i] - 0x30)) % mod; 
    for (int i = 0; i < sb.size(); i++) b = (b * 10 + (sb[i] - 0x30)) % mod; 
    return powLL(a, b); 
} 
 
  
 
int main(void) 
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    string sa, sb; cin >> sa >> sb;
    cout << powStrings(sa, sb) << endl; 
    return EXIT_SUCCESS; 
}
    • 1

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