计算(a * b)％c，使(a％c)*(b％c)可以超出范围

给定三个数字a，b和c，使得a，b和c最多为10 ¹⁶ 。任务是计算(a * b)％c
一个简单的((a％c)*(b％c))％c的解决方案在这里不起作用。这里的问题是a和b可能很大，因此当我们计算(a％c)*(b％c)时，它超出了long long int可以容纳的范围，因此会发生溢出。例如，如果a =(10 ¹² +7)，b =(10 ¹³ +5)，c =(10 ¹⁵ +3)。
现在long long int最多可以容纳4 x 10 ¹⁸ (大约)，并且a * b远大于该值。

代替进行直接乘法，我们可以添加find a + a +………。(b次)，并在每次添加a时取c模，这样就不会发生溢出。但是，从约束a，b和c的角度来看，这将是低效的。我们必须以某种方式以优化的方式计算(∑ a)％c。
我们可以使用分治法来计算它。主要思想是：

1. 如果b是偶数，则a * b =(2 * a)*(b / 2)
2. 如果b为奇数，则a * b = a +(2 * a)*((b-1)/ 2)

下面是该算法的实现：

C++

// C++ program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
#include 
using namespace std;
typedef long long int ll;
 
// returns (a*b)%c
ll mulmod(ll a,ll b,ll c)
{
    // base case if b==0, return 0
    if (b==0)
        return 0;
 
    // Divide the problem into 2 parts
    ll s = mulmod(a, b/2, c);
 
    // If b is odd, return
    // (a+(2*a)*((b-1)/2))%c
    if (b%2==1)
        return (a%c+2*(s%c)) % c;
 
    // If b is odd, return
    // ((2*a)*(b/2))%c
    else
        return (2*(s%c)) % c;
}
 
// Driver code
int main()
{
    ll a = 1000000000007, b = 10000000000005;
    ll c = 1000000000000003;
    printf("%lldn", mulmod(a, b, c));
    return 0;
}

Java

// Java program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
// returns (a*b)%c
class GFG {
 
    static long mulmod(long a, long b, long c) {
        // base case if b==0, return 0
        if (b == 0) {
            return 0;
        }
 
        // Divide the problem into 2 parts
        long s = mulmod(a, b / 2, c);
 
        // If b is odd, return
        // (a+(2*a)*((b-1)/2))%c
        if (b % 2 == 1) {
            return (a % c + 2 * (s % c)) % c;
        } // If b is odd, return
        // ((2*a)*(b/2))%c
        else {
            return (2 * (s % c)) % c;
        }
    }
 
// Driver code
    public static void main(String[] args) {
        long a = 1000000000007L, b = 10000000000005L;
        long c = 1000000000000003L;
        System.out.println((mulmod(a, b, c)));
    }
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program of above approach
 
# returns (a*b)%c
def mulmod(a, b, c):
 
    # base case if b==0, return 0
    if(b == 0):
        return 0
 
    # Divide the problem into 2 parts
    s = mulmod(a, b // 2, c)
 
    # If b is odd, return
    # (a+(2*a)*((b-1)/2))%c
    if(b % 2 == 1):
        return (a % c + 2 * (s % c)) % c
 
    # If b is odd, return
    # ((2*a)*(b/2))%c
    else:
        return (2 * (s % c)) % c
 
# Driver code
if __name__=='__main__':
    a = 1000000000007
    b = 10000000000005
    c = 1000000000000003
    print(mulmod(a, b, c))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
using System;
 
// returns (a*b)%c
class GFG
{
static long mulmod(long a, long b, long c)
{
    // base case if b==0, return 0
    if (b == 0)
        return 0;
 
    // Divide the problem into 2 parts
    long s = mulmod(a, b / 2, c);
 
    // If b is odd, return
    // (a+(2*a)*((b-1)/2))%c
    if (b % 2 == 1)
        return (a % c + 2 * (s % c)) % c;
 
    // If b is odd, return
    // ((2*a)*(b/2))%c
    else
        return (2 * (s % c)) % c;
}
 
// Driver code
public static void Main()
{
    long a = 1000000000007, b = 10000000000005;
    long c = 1000000000000003;
    Console.WriteLine(mulmod(a, b, c));
}
}
 
// This code is contributed by mits

Javascript

 
// javascript program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
// returns (a*b)%c  
function mulmod(a , b , c) {
 
        // base case if b==0, return 0
        if (b == 0) {
            return 0;
        }
 
        // Divide the problem into 2 parts
        var s = mulmod(a, parseInt(b / 2), c);
 
        // If b is odd, return
        // (a+(2*a)*((b-1)/2))%c
        if (b % 2 == 1) {
            return (a % c + 2 * (s % c)) % c;
        }
        // If b is odd, return
        // ((2*a)*(b/2))%c
        else {
            return (2 * (s % c)) % c;
        }
    }
 
// Driver code
var a = 1000000000007, b = 10000000000005;
var c = 1000000000000003;
document.write((mulmod(a, b, c)));
 
// This code contributed by Princi Singh
 

输出 ：

74970000000035