Is there any quick way to determine first k digits on n^n

I am writing a program where I need to know only the first k (k can be anywhere between 1-5) numbers of another big number which can be represented as n^n where n is a very large number.

Currently I am actually calculating n^n and then parsing it as a string.开发者_JAVA技巧 I wonder if there is a better more fast method exists.

There are two possibilities.

If you want the first k leading digits (as in: the leading digit of 12345 is 1), then you can use the fact that

n^n = 10^(n*Log10(n))

so you compute the fractional part f of n*Log10(n), and then the first k digits of 10^f will be your result. This works for numbers up to about 10^10 before round-off errors start kicking in if you use double precision. For example, for n = 2^20, f = 0.57466709..., 10^f = 3.755494... so your first 5 digits are 37554. For n = 4, f = 0.4082..., 10^f = 2.56 so your first digit is 2.

If you want the first k trailing digits (as in: the trailing digit of 12345 is 5), then you can use modular arithmetic. I would use the squaring trick:

factor = n mod 10^k
result = 1
while (n != 0) 
    if (n is odd) then result = (result * factor) mod 10^k
    factor = (factor * factor) mod 10^k
    n >>= 1

Taking n=2^20 as an example again, we find that result = 88576. For n=4, we have factor = 1, 4, 6 and result = 1, 1, 6 so the answer is 6.

if you mean the least significant or rightmost digits, this can be done with modular multiplication. It's O(N) complexity and doesn't require any special bignum data types.

#include <cmath>
#include <cstdio>

//returns ((base ^ exponent) % mod)
int modularExponentiation(int base, int exponent, int mod){
    int result = 1;
    for(int i = 0; i < exponent; i++){
        result = (result * base) % mod;
    }
    return result;
}

int firstKDigitsOfNToThePowerOfN(int k, int n){
    return modularExponentiation(n, n, pow(10, k));
}

int main(){
    int n = 11;
    int result = firstKDigitsOfNToThePowerOfN(3, n);
    printf("%d", result); 
}

This will print 611, the first three digits of 11^11 = 285311670611.

This implementation is suitable for values of N less than sqrt(INT_MAX), which will vary but on my machine and language it's over 46,000.

Furthermore, if it so happens that your INT_MAX is less than (10^k)^2, you can change modularExponentiation to handle any N that can fit in an int:

int modularExponentiation(int base, int exponent, int mod){
    int result = 1;
    for(int i = 0; i < exponent; i++){
        result = (result * (base % mod)) % mod; //doesn't overflow as long as mod * mod < INT_MAX
    }
    return result;
}

if O(n) time is insufficient for you, we can take advantage of the property of exponentiation that A^(2*C) = (A^C)^2, and get logarithmic efficiency.

//returns ((base ^ exponent) % mod)
int modularExponentiation(int base, int exponent, int mod){
    if (exponent == 0){return 1;}
    if (exponent == 1){return base % mod;}
    if (exponent % 2 == 1){
        return ((base % mod) * modularExponentiation(base, exponent-1, mod)) % mod;
    }
    else{
        int newBase = modularExponentiation(base, exponent / 2, mod);
        return (newBase * newBase) % mod;
    }
}