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Calculating (a power b) mod p
By devarshi09, history, 6 years ago, In English

How to calculate (a power b) mod p correctly if b is exceeding long long integer limit in c++.

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6 years ago, # |
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O(log2(b))?

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6 years ago, # |
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Let b = b1b2.....bn(concatenated) By induction, Let x = a ^ (b1b2...bk) (mod p)

a ^ (b1b2....b(k+1)) = ( a^(b1b2....bk) ) ^ 10 * (a ^ b(k+1)) = x^10 * a^b(k+1)

now you can calculate modulo without overflow

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6 years ago, # |
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first i need to store b. But b is exceeding the integer limit! How can i store b so that i correctly calculate (a power b) mod p. p is prime.

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    6 years ago, # ^ |
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    If b is in the form of a string, then we can calculate with the following code:

    ll B = 0;
for (char c: b) {
    B = (10*B+(c-'0'))%(p-1);
}

    Now assuming that a is nonzero.

    EDIT: This is exactly what DongwonShin wrote above. If b is being calculated via a DP table, then you can just take all values mod p - 1 and there won't be overflow.

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      6 years ago, # ^ |
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      got it. thanks!

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      6 years ago, # ^ |
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      but why p-1?

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        6 years ago, # ^ |
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        Fermat's theorem tells us that if (a, p) = 1. So we can take the power by the modulo.

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          6 years ago, # ^ |
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          Thanks man!

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            6 years ago, # ^ |
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            In general if you have two co prime integers m and n, then m^p modulo n = m^(p%f(n)) modulo n, where f(n) is Euler totient function This.

            For any prime n, f(n)=n-1.So, the Fermat's theorem follows from this.

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              6 years ago, # ^ |
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              I am using precomputed factorial and inverse factorial array to calculate b so should i use n-1 as the mod there also?

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      6 years ago, # ^ |
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      What if b is being calculated through nCr using precomputed factorial and inverse modulo arrays?Then also we've to use same mod p-1 ?

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        6 years ago, # ^ |
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        As long as your base(b) and p are co prime, I don't see why it can't be done. So, yes use same mod (p-1).

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          6 years ago, # ^ |
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          I tried but then for example 11*(inv[11)) didn't turn out to be 1 when i used p-1 as mod but while using p as mod it did show 1. here inv[11] = power(11,mod-2,mod) where power is fast exponential function.

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      6 years ago, # ^ |
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      What if b is calculated as a-c and c requires division operation?Then while using inverse modulo for division i should use which mod?I am confused.Please help.

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6 years ago, # |
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Refer https://www.geeksforgeeks.org/find-abm-where-b-is-very-large/

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