添加负数和正数最多 10^100000

Question

I've been trying to solve this problem (from school) for just about a week now. We're given two numbers, from -(10^100000) to +that.

Of course the simplest solution is to implement written addition, so that's what I did. I decided, that I would store the numbers as strings, using two functions:

int ti(char a) { // changes char to int
    int output = a - 48;
    return output;
}

char tc(int a) { // changes int to char
    char output = a + 48;
    return output;
}

This way I can store negative digits, like -2. With that in mind I implemented a toMinus function:

void toMinus(std::string &a) { // 123 -> -1 -2 -3
    for (auto &x : a) {
        x = tc(-ti(x));
    }
}

I also created a changeSize function, which adds 0 to the beginning of the number until they are both their max size + 1 and removeZeros, which removes leading zeros:

void changeSize(std::string &a, std::string &b) {
    size_t exp_size = std::max(a.size(), b.size()) + 2;
    while (a.size() != exp_size) {
        a = '0' + a;
    }

    while (b.size() != exp_size) {
        b = '0' + b;
    }
}

void removeZeros(std::string &a) {
    int i = 0;    
    for (; i < a.size(); i++) {
        if (a[i] != '0') {
            break;
        }
    }
    a.erase(0, i);
    if (a.size() == 0) {
        a = "0";
    }
}

After all that, I created the main add() function:

std::string add(std::string &a, std::string &b) {
    bool neg[2] = {false, false};
    bool out_negative = false;
    if (a[0] == '-') {
        neg[0] = true;
        a.erase(0, 1);
    }
    if (b[0] == '-') {
        neg[1] = true;
        b.erase(0, 1);
    }

    changeSize(a, b);

    if (neg[0] && !(neg[1] && neg[0])) {
        toMinus(a);
    }
    if(neg[1] && !(neg[1] && neg[0])) {
        toMinus(b);
    }

    if (neg[1] && neg[0]) {
        out_negative = true;
    }

    // Addition
    for (int i = a.size() - 1; i > 0; i--) {
        int _a = ti(a[i]);
        int _b = ti(b[i]);
        int out = _a + _b;
        if (out >= 10) {
            a[i - 1] += out / 10;
        } else if (out < 0) {
            if (abs(out) < 10) {
                a[i - 1]--;
            } else {
                a[i - 1] += abs(out) / 10;
            }
            if (i != 1)
                out += 10;
        }

        a[i] = tc(abs(out % 10));
    }

    if (ti(a[0]) == -1) { // Overflow
        out_negative = true;
        a[0] = '0';
        a[1]--;

        for (int i = 2; i < a.size(); i++) {
            if (i == a.size() - 1) {
                a[i] = tc(10 - ti(a[i]));
            } else {
                a[i] = tc(9 - ti(a[i]));
            }
        }
    }

    if (neg[0] && neg[1]) {
        out_negative = true;
    }

    removeZeros(a);

    if (out_negative) {
        a = '-' + a;
    }

    return a;
}

This program works in most cases, although our school checker found that it doesn't - like instead of

-4400547114413430129608370706728634555709161366260921095898099024156859909714382493551072616612065064

it returned

-4400547114413430129608370706728634555709161366260921095698099024156859909714382493551072616612065064

I can't find what the problem is. Please help and thank you in advance.

Full code on pastebin

Answer 1

While I think your overall approach is totally reasonable for this problem, your implementation seems a bit too complicated. Trying to solve this myself, I came up with this:

#include <iostream>
#include <limits>
#include <random>
#include <string>

bool greater(const std::string& a, const std::string& b)
{
    if (a.length() == b.length()) return a > b;
    return a.length() > b.length();
}

std::string add(std::string a, std::string b)
{
    std::string out;

    bool aNeg = a[0] == '-';
    if (aNeg) a.erase(0, 1);
    bool bNeg = b[0] == '-';
    if (bNeg) b.erase(0, 1);

    bool resNeg = aNeg && bNeg;
    if (aNeg ^ bNeg && (aNeg && greater(a, b) || bNeg && greater(b, a)))
    {
        resNeg = true;
        std::swap(a, b);
    }

    int i = a.length() - 1;
    int j = b.length() - 1;
    int carry = 0;
    while (i >= 0 || j >= 0)
    {
        const int digitA = (i >= 0) ? a[i] - '0' : 0;
        const int digitB = (j >= 0) ? b[j] - '0' : 0;
        const int sum = (aNeg == bNeg ? digitA + digitB : (bNeg ? digitA - digitB : digitB - digitA)) + carry;
        carry = 0;
        if (sum >= 10) carry = 1;
        else if (sum < 0) carry = -1;

        out = std::to_string((sum + 20) % 10) + out;

        i--;
        j--;
    }
    if (carry) out = '1' + out;
    while (out[0] == '0') out.erase(0, 1);
    if (resNeg) out = '-' + out;
    return out;
}


void test()
{
    std::random_device rd;
    std::mt19937 gen(rd());
    std::uniform_int_distribution<> dis(-std::numeric_limits<int32_t>::max(), std::numeric_limits<int32_t>::max());
    for (int i = 0; i < 1000000; ++i)
    {
        const int64_t a = dis(gen);
        const int64_t b = dis(gen);
        const auto expected = std::to_string(a + b);
        const auto actual = add(std::to_string(a), std::to_string(b));
        if (actual != expected) {
            std::cout << "mismatch expected: " << expected << std::endl;
            std::cout << "mismatch actual  : " << actual << std::endl;
            std::cout << "                a: " << a << std::endl;
            std::cout << "                b: " << b << std::endl;
        }
    }
}

int main()
{
    test();
}

It can potentially be further optimized, but the main points are:

• If the sign of both numbers is the same, we can do simple written addition. If both are negative, we simply prepend - at the end.
• If the signs are different, we do written subtraction. If the minuend is greater than the subtrahend, there's no issue, we know that the result will be positive. If, however, the subtrahend is greater, we have to reformulate the problem. For example, 123 - 234 we would formulate as -(234 - 123) . The inner part we can solve using regular written subtraction, after which we prepend - .

I test this with random numbers for which we can calculate the correct result using regular integer arithmetic. Since it doesn't fail for those, I'm pretty confident it also works correctly for larger inputs. An approach like this could also help you uncover cases where your implementation fails.

Other than that, I think you should use a known failing case with a debugger or simply print statements for the intermediate steps to see where it fails. The only small differences in the failing example you posted could point at some issue with handling a carry-over.