來自字符串的C ++整數

Question

如何將std::string轉換為無符號整數，並允許非常長的輸入？

例如，輸入5000000000應返回705032704 （ 5000000000 mod 2^32 ），假設此處無符號為32位。 輸入9999999999999999999999999999應返回268435455 。

std::stoi和朋友給出std::out_of_range時提供了這樣一個龐大的數字。

在給定此類輸入的情況下，使用std::istringstream::operator>>(unsigned)失敗。

是否有任何函數將字符串轉換為整數，而不是在大輸入的情況下挽救？ （如果可能的話，我寧願避免自己寫一個。）

Answer 1

你可以自己編寫一個函數：

unsigned int get_uint(const std::string &s) {
    unsigned int r = 0U;
    for(auto c : s) {
        assert(std::isdigit(c));
        r = r * 10 + (c - '0');
    }
    return r;
}

實例

這是有效的，因為無符號溢出在C ++中用作模運算。

從3.9.1 / 4開始

聲明無符號的無符號整數應遵守算術模2 ^ n的定律，其中n是特定整數大小的值表示中的位數

Answer 2

與Python和Java不同，C ++中沒有對bigint的標准支持。 但是有許多庫可以為您提供所需的支持。 這是我在C ++中使用bigint的一個實現，我有時會在競爭性編程中使用它：

#include <vector>
#include <cstdlib>
#include <iostream>
#include <iomanip>
#include <string>
#include <ctime>
using namespace std;

const int base = 1000000000;
const int base_digits = 9;

struct bigint
{
vector<int> a;
int sign;

bigint() :
    sign(1)
{
}

bigint(long long v)
{
    *this = v;
}

bigint(const string &s)
{
    read(s);
}

void operator=(const bigint &v)
{
    sign = v.sign;
    a = v.a;
}

void operator=(long long v)
{
    sign = 1;
    if (v < 0)
        sign = -1, v = -v;
    for (; v > 0; v = v / base)
        a.push_back(v % base);
}

bigint operator+(const bigint &v) const
{
    if (sign == v.sign)
    {
        bigint res = v;

        for (int i = 0, carry = 0; i < (int) max(a.size(), v.a.size()) || carry; ++i)
        {
            if (i == (int) res.a.size())
                res.a.push_back(0);
            res.a[i] += carry + (i < (int) a.size() ? a[i] : 0);
            carry = res.a[i] >= base;
            if (carry)
                res.a[i] -= base;
        }
        return res;
    }
    return *this - (-v);
}

bigint operator-(const bigint &v) const
{
    if (sign == v.sign)
    {
        if (abs() >= v.abs())
        {
            bigint res = *this;
            for (int i = 0, carry = 0; i < (int) v.a.size() || carry; ++i)
            {
                res.a[i] -= carry + (i < (int) v.a.size() ? v.a[i] : 0);
                carry = res.a[i] < 0;
                if (carry)
                    res.a[i] += base;
            }
            res.trim();
            return res;
        }
        return -(v - *this);
    }
    return *this + (-v);
}

void operator*=(int v)
{
    if (v < 0)
        sign = -sign, v = -v;
    for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i)
    {
        if (i == (int) a.size())
            a.push_back(0);
        long long cur = a[i] * (long long) v + carry;
        carry = (int) (cur / base);
        a[i] = (int) (cur % base);
    }
    trim();
}

bigint operator*(int v) const
{
    bigint res = *this;
    res *= v;
    return res;
}

friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1)
{
    int norm = base / (b1.a.back() + 1);
    bigint a = a1.abs() * norm;
    bigint b = b1.abs() * norm;
    bigint q, r;
    q.a.resize(a.a.size());

    for (int i = a.a.size() - 1; i >= 0; i--)
    {
        r *= base;
        r += a.a[i];
        int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
        int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
        int d = ((long long) base * s1 + s2) / b.a.back();
        r -= b * d;
        while (r < 0)
            r += b, --d;
        q.a[i] = d;
    }

    q.sign = a1.sign * b1.sign;
    r.sign = a1.sign;
    q.trim();
    r.trim();
    return make_pair(q, r / norm);
}

bigint operator/(const bigint &v) const
{
    return divmod(*this, v).first;
}

bigint operator%(const bigint &v) const
{
    return divmod(*this, v).second;
}

void operator/=(int v)
{
    if (v < 0)
        sign = -sign, v = -v;
    for (int i = (int) a.size() - 1, rem = 0; i >= 0; --i)
    {
        long long cur = a[i] + rem * (long long) base;
        a[i] = (int) (cur / v);
        rem = (int) (cur % v);
    }
    trim();
}

bigint operator/(int v) const
{
    bigint res = *this;
    res /= v;
    return res;
}

int operator%(int v) const
{
    if (v < 0)
        v = -v;
    int m = 0;
    for (int i = a.size() - 1; i >= 0; --i)
        m = (a[i] + m * (long long) base) % v;
    return m * sign;
}

void operator+=(const bigint &v)
{
    *this = *this + v;
}
void operator-=(const bigint &v)
{
    *this = *this - v;
}
void operator*=(const bigint &v)
{
    *this = *this * v;
}
void operator/=(const bigint &v)
{
    *this = *this / v;
}

bool operator<(const bigint &v) const
{
    if (sign != v.sign)
        return sign < v.sign;
    if (a.size() != v.a.size())
        return a.size() * sign < v.a.size() * v.sign;
    for (int i = a.size() - 1; i >= 0; i--)
        if (a[i] != v.a[i])
            return a[i] * sign < v.a[i] * sign;
    return false;
}

bool operator>(const bigint &v) const
{
    return v < *this;
}
bool operator<=(const bigint &v) const
{
    return !(v < *this);
}
bool operator>=(const bigint &v) const
{
    return !(*this < v);
}
bool operator==(const bigint &v) const
{
    return !(*this < v) && !(v < *this);
}
bool operator!=(const bigint &v) const
{
    return *this < v || v < *this;
}

void trim()
{
    while (!a.empty() && !a.back())
        a.pop_back();
    if (a.empty())
        sign = 1;
}

bool isZero() const
{
    return a.empty() || (a.size() == 1 && !a[0]);
}

bigint operator-() const
{
    bigint res = *this;
    res.sign = -sign;
    return res;
}

bigint abs() const
{
    bigint res = *this;
    res.sign *= res.sign;
    return res;
}

long long longValue() const
{
    long long res = 0;
    for (int i = a.size() - 1; i >= 0; i--)
        res = res * base + a[i];
    return res * sign;
}

friend bigint gcd(const bigint &a, const bigint &b)
{
    return b.isZero() ? a : gcd(b, a % b);
}
friend bigint lcm(const bigint &a, const bigint &b)
{
    return a / gcd(a, b) * b;
}

void read(const string &s)
{
    sign = 1;
    a.clear();
    int pos = 0;
    while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+'))
    {
        if (s[pos] == '-')
            sign = -sign;
        ++pos;
    }
    for (int i = s.size() - 1; i >= pos; i -= base_digits)
    {
        int x = 0;
        for (int j = max(pos, i - base_digits + 1); j <= i; j++)
            x = x * 10 + s[j] - '0';
        a.push_back(x);
    }
    trim();
}

friend istream& operator>>(istream &stream, bigint &v)
{
    string s;
    stream >> s;
    v.read(s);
    return stream;
}

friend ostream& operator<<(ostream &stream, const bigint &v)
{
    if (v.sign == -1)
        stream << '-';
    stream << (v.a.empty() ? 0 : v.a.back());
    for (int i = (int) v.a.size() - 2; i >= 0; --i)
        stream << setw(base_digits) << setfill('0') << v.a[i];
    return stream;
}

static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits)
{
    vector<long long> p(max(old_digits, new_digits) + 1);
    p[0] = 1;
    for (int i = 1; i < (int) p.size(); i++)
        p[i] = p[i - 1] * 10;
    vector<int> res;
    long long cur = 0;
    int cur_digits = 0;
    for (int i = 0; i < (int) a.size(); i++)
    {
        cur += a[i] * p[cur_digits];
        cur_digits += old_digits;
        while (cur_digits >= new_digits)
        {
            res.push_back(int(cur % p[new_digits]));
            cur /= p[new_digits];
            cur_digits -= new_digits;
        }
    }
    res.push_back((int) cur);
    while (!res.empty() && !res.back())
        res.pop_back();
    return res;
}

typedef vector<long long> vll;

static vll karatsubaMultiply(const vll &a, const vll &b)
{
    int n = a.size();
    vll res(n + n);
    if (n <= 32)
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                res[i + j] += a[i] * b[j];
        return res;
    }

    int k = n >> 1;
    vll a1(a.begin(), a.begin() + k);
    vll a2(a.begin() + k, a.end());
    vll b1(b.begin(), b.begin() + k);
    vll b2(b.begin() + k, b.end());

    vll a1b1 = karatsubaMultiply(a1, b1);
    vll a2b2 = karatsubaMultiply(a2, b2);

    for (int i = 0; i < k; i++)
        a2[i] += a1[i];
    for (int i = 0; i < k; i++)
        b2[i] += b1[i];

    vll r = karatsubaMultiply(a2, b2);
    for (int i = 0; i < (int) a1b1.size(); i++)
        r[i] -= a1b1[i];
    for (int i = 0; i < (int) a2b2.size(); i++)
        r[i] -= a2b2[i];

    for (int i = 0; i < (int) r.size(); i++)
        res[i + k] += r[i];
    for (int i = 0; i < (int) a1b1.size(); i++)
        res[i] += a1b1[i];
    for (int i = 0; i < (int) a2b2.size(); i++)
        res[i + n] += a2b2[i];
    return res;
}

bigint operator*(const bigint &v) const
{
    vector<int> a6 = convert_base(this->a, base_digits, 6);
    vector<int> b6 = convert_base(v.a, base_digits, 6);
    vll a(a6.begin(), a6.end());
    vll b(b6.begin(), b6.end());
    while (a.size() < b.size())
        a.push_back(0);
    while (b.size() < a.size())
        b.push_back(0);
    while (a.size() & (a.size() - 1))
        a.push_back(0), b.push_back(0);
    vll c = karatsubaMultiply(a, b);
    bigint res;
    res.sign = sign * v.sign;
    for (int i = 0, carry = 0; i < (int) c.size(); i++)
    {
        long long cur = c[i] + carry;
        res.a.push_back((int) (cur % 1000000));
        carry = (int) (cur / 1000000);
    }
    res.a = convert_base(res.a, 6, base_digits);
    res.trim();
    return res;
}
};

int main()
{
bigint     a("99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999");
bigint b("19999999999999999999999999999999999999999999999999999999999999999999999999999999999999998");
cout << a * b << endl;
cout << a / b << endl;

string sa, sb;
for (int i = 0; i < 100000; i++)
    sa += i % 10 + '0';
for (int i = 0; i < 20000; i++)
    sb += i % 10 + '0';
a = bigint(sa);
b = bigint(sb);

clock_t start = clock();
bigint c = a / b;
fprintf(stderr, "time=%.3lfsec\n", 0.001 * (clock() - start));
return 0;
}

Answer 3

如果適合您的范圍，請使用std::stoull ，否則請使用C ++中的BigInteger庫

unsigned long long stoull(const std::string & str, std::size_t * pos = 0, int base = 10);

一個不錯的bigint庫是https://mattmccutchen.net/bigint/ 。