# How to avoid overflow in modular multiplication?

Consider below simple method to multiply two numbers.

```// A Simple solution that causes overflow when
// value of (a % mod) * (b % mod) becomes more than
// maximum value of long long int
#define ll long long

ll multiply(ll a, ll b, ll mod)
{
return ((a % mod) * (b % mod)) % mod;
}
```

The above function works fine when multiplication doesn’t result in overflow. But if input numbers are such that the result of multiplication is more than maximum limit.

For example, the above method fails when mod = 1011, a = 9223372036854775807 (largest long long int) and b = 9223372036854775807 (largest long long int). Note that there can be smaller values for which it may fail. There can be many more examples of smaller values. In fact any set of values for which multiplication can cause a value greater than maximum limit.

How to avoid overflow?
We can multiply recursively to overcome the difficulty of overflow. To multiply a*b, first calculate a*b/2 then add it twice. For calculating a*b/2 calculate a*b/4 and so on (similar to log n exponentiation algorithm).

```// To compute (a * b) % mod
multiply(a,  b, mod)
1)  ll res = 0; // Initialize result
2)  a = a % mod.
3)  While (b > 0)
a) If b is odd, then add 'a' to result.
res = (res + a) % mod
b) Multiply 'a' with 2
a = (a * 2) % mod
c) Divide 'b' by 2
b = b/2
4)  Return res ```

Below is C++ implementation.

```// C++ program for modular multiplication without
// any overflow
#include<iostream>
using namespace std;

typedef long long int ll;

// To compute (a * b) % mod
ll mulmod(ll a, ll b, ll mod)
{
ll res = 0; // Initialize result
a = a % mod;
while (b > 0)
{
// If b is odd, add 'a' to result
if (b % 2 == 1)
res = (res + a) % mod;

// Multiply 'a' with 2
a = (a * 2) % mod;

// Divide b by 2
b /= 2;
}

// Return result
return res % mod;
}

// Driver program
int main()
{
ll a = 9223372036854775807, b = 9223372036854775807;
cout << mulmod(a, b, 100000000000);
return 0;
}
```

Output:

`84232501249`

Thanks to Utkarsh Trivedi for suggesting above solution.

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