找出分数 a/b 的小数点后第 k 位，其中 a,b,k 是非常大的整数（小于 10e18）

Question

I was tasked to find a digit of kth position after the decimal point of a fraction(a/b). Yesterday I found out this algorithm.
To get any digit after the decimal point, I generate a variable called rem and make a loop

for (int i = 1; i <= k+1; i++)    
      {
         rem = a%b;
         a = rem*10;
      }
      cout << a/b;    

the loop will return a value being the kth digit after the decimal point.
However the task require me to calculate with a,b,k being very large number ( less or equal to 10e18), so it's sure that the code will exceed the time limit.

• Find the number of the digits before the repetend. It's the greater of the numbers of 2 and 5 factors in the denominator.
• If the k doesn't exceed the number of digits, run the for loop.
• Else, we will still run the for loop to k+1. Store the value of the remainder of the division in a variable x.
• Run a while loop with the same content above until the remainder again have the value of x. Since then, store every quotient of the division into an array name qut.
• After the while loop terminates, the array will have stored every digit inside the repetend. According to the number of the digits inside the array, we can calculate the kth digit.
  However, this algorithm still prove to be time consuming because in the case of a and b are two consecutive integer, the repetend become very large. Can you help me with any idea?

Answer 1

What your for loop calculates is effectively 10 * (a*10 ^k % b) / b. We can do this more efficiently by doing exponentiation by squaring. We have to be careful not to overflow at every point though:

int kth_digit_frac(uint64_t a, uint64_t b, uint64_t k) {
    return 10 * mulmodu64(a, powmod(10, k, b), b) / b;
}

// a*b % m, overflow safe
inline uint64_t mulmodu64(uint64_t a, uint64_t b, uint64_t m) {
    #if defined(__GNUC__) && defined(__x86_64__)
        uint64_t q, r;
        asm("mulq %3;"
            "divq %4;"
            : "=a"(q), "=d"(r)
            : "a"(a), "d"(b), "rm"(m)
            : "cc");
        return r;
    #else
        a %= m;
        b %= m;

        // No overflow possible.
        if (a == 0) return 0;
        if (b <= std::numeric_limits<uint64_t>::max() / a) return (a*b) % m;

        uint64_t res = 0;
        while (a != 0) {
            if (a & 1) {
                if (b >= m - res) res -= m;
                res += b;
            }

            a >>= 1;
            if (b >= m - b) b += b - m;
            else            b += b;
        }

        return res;
    #endif
}


// b^e % m, overflow safe
inline uint64_t powmod(uint64_t b, uint64_t e, uint64_t m) {
    uint64_t r = 1;

    b %= m;
    while (e) {
        if (e % 2 == 1) r = mulmodu64(r, b, m);
        b = mulmodu64(b, b, m);
        e >>= 1;
    }

    return r;
}

It runs in the blink of an eye for any a,b,k that fit in 64 bit integers.

Answer 2

I believe two of the solutions above is still incorrect, because of a little error.

I think the following will work nicely:

10 * (a*10^(k-1) % b) / b