给定两个整数A和B ,其中A不能被B整除,任务是将A / B表示为自然数模m ,其中m = 1000000007 。
注意:在需要表达事件的概率,曲线面积和多边形等的情况下,此表示很有用。
例子:
Input: A = 2, B = 6
Output: 333333336
Input: A = 4, B = 5
Output: 600000005
方法:我们知道, A / B可以写成A *(1 / B),即A *(B ^ -1) 。
已知模(%)运算符满足以下关系:
(a * b) % m = ( (a % m) * (b % m) ) % m
因此,我们可以这样写:
(b ^ -1) % m = (b ^ m-2) % m (Fermat's little theorem)
因此结果将是:
( (A mod m) * ( power(B, m-2) % m) ) % m
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long int
#define m 1000000007
// Function to return the GCD of given numbers
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Recursive function to return (x ^ n) % m
ll modexp(ll x, ll n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
ll getFractionModulo(ll a, ll b)
{
ll c = gcd(a, b);
a = a / c;
b = b / c;
// (b ^ m-2) % m
ll d = modexp(b, m - 2);
// Final answer
ll ans = ((a % m) * (d % m)) % m;
return ans;
}
// Driver code
int main()
{
ll a = 2, b = 6;
cout << getFractionModulo(a, b) << endl;
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG {
static long m = 1000000007;
// Function to return the GCD of given numbers
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
static long getFractionModulo(long a, long b)
{
long c = gcd(a, b);
a = a / c;
b = b / c;
// (b ^ m-2) % m
long d = modexp(b, m - 2);
// Final answer
long ans = ((a % m) * (d % m)) % m;
return ans;
}
// Driver code
public static void main (String[] args) {
long a = 2, b = 6;
System.out.println(getFractionModulo(a, b));
}
}
// This code is contributed by inder_verma
Python3
# Python3 implementation of the approach
m = 1000000007
# Function to return the GCD
# of given numbers
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
# Recursive function to return (x ^ n) % m
def modexp(x, n):
if (n == 0) :
return 1
elif (n % 2 == 0) :
return modexp((x * x) % m, n // 2)
else :
return (x * modexp((x * x) % m,
(n - 1) / 2) % m)
# Function to return the fraction modulo mod
def getFractionModulo(a, b):
c = gcd(a, b)
a = a // c
b = b // c
# (b ^ m-2) % m
d = modexp(b, m - 2)
# Final answer
ans = ((a % m) * (d % m)) % m
return ans
# Driver code
if __name__ == "__main__":
a = 2
b = 6
print ( getFractionModulo(a, b))
# This code is contributed by ita_c
C#
//C# implementation of the approach
using System;
public class GFG{
static long m = 1000000007;
// Function to return the GCD of given numbers
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
static long getFractionModulo(long a, long b)
{
long c = gcd(a, b);
a = a / c;
b = b / c;
// (b ^ m-2) % m
long d = modexp(b, m - 2);
// Final answer
long ans = ((a % m) * (d % m)) % m;
return ans;
}
// Driver code
static public void Main (){
long a = 2, b = 6;
Console.WriteLine(getFractionModulo(a, b));
}
}
PHP
输出:
333333336
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