Question

如何在变量中存储一个大数字并使用for循环？
我有一个非常大的数字75472202764752234070123900087933251，我需要从0循环到这个数字！
甚至可以这样做吗？结束需要多长时间？

编辑：我正试图用蛮力解决一个难题。它是一个组合问题。强制案件可能达到470C450 所以我想我应该使用不同的算法...

Answer 1

这可能需要如果C ++每秒处理100,000个循环，则为0.23 x 10 ^ 23年：|

http://www.wolframalpha.com/input/?i=75472202764752234070123900087933251%2F%28100000*1*3600*24*365%29

Answer 2

好吧，你需要一个能够在非常大的整数上处理至少一部分初始化，布尔和算术函数的实现。类似于：https://mattmccutchen.net/bigint/。

对于能够提供比一般大整数数学库更好的性能的东西，您可以专门使用专门的操作来允许使用大整数作为计数器。有关此示例，请参阅dewtell's updated answer此问题。

至于你可以从0循环到那个数字：嗯，是的，有可能用上面的解决方案之一为它编写代码，但我认为答案是否定的，你个人不会能够做到这一点，因为看到它完成后你将不会活着。

[编辑：是的，我肯定会建议您找到不同的算法。：d]

Answer 3

看起来这个数字符合128位。因此，您可以使用现代系统和实现此类数字的现代编译器。例如，使用gcc作为编译器的64位linux系统就是这种情况。这类似__uint128_t，你可以使用。

显然你不能将这样的变量用作for - 循环变量，其他人已经给你计算了。但你可以用它来存储你的一些计算。

Answer 4

如果你需要循环一定次数，并且该数字大于2 ^ 64，那么只需使用while(1)，因为无论如何你的计算机在最多计数到2 ^ 64之前都会中断。

Answer 5

不需要完整的bignum包 - 如果您只需要一个循环计数器，这里是一个简单的字节计数器，它使用一个字节数组作为计数器。当字节数组再次包围所有零时它停止。如果你想要计算除2 ^（bytesUsed * CHAR_BITS）之外的其他值，你可以计算你想要的迭代次数的负数的二进制补码值，并让它计数到0，记住字节数[0]是低位字节（或使用正值并向下计数而不是向上）。

#include <stdio.h>
#define MAXBYTES 20
/* Simple byte counter - note it uses argc as # of bytes to use for convenience */
int main(int argc, char **argv) {
    unsigned char bytes[MAXBYTES];
    const int bytesUsed = argc < MAXBYTES? argc : MAXBYTES;
    int i;
    unsigned long counter = (unsigned long)-1; /* to give loop something to do */
    for (i = 0; i < bytesUsed; i++) bytes[i] = 0; /* Initialize bytes */
    do {
        for (i = 0; i < bytesUsed && !++bytes[i]; i++) ; /* NULL BODY - this is the byte counter */
        counter++;
    } while (i < bytesUsed);
    printf("With %d bytes used, final counter value = %lu\n", bytesUsed, counter);
}

前4个值的运行时间（在Cygwin下，在Lenovo T61上）：

$ time ./bytecounter

使用1个字节，最终计数器值= 255

真正的0m0.078s 用户0m0.031s sys 0m0.046s

$ time ./bytecounter a

使用2个字节，最终计数器值= 65535

真正的0m0.063s 用户0m0.031s sys 0m0.031s

$ time ./bytecounter a a

使用3个字节，最终计数器值= 16777215

真正的0m0.125s 用户0m0.015s sys 0m0.046s

$ time ./bytecounter a a a

使用4个字节，最终计数器值= 4294967295

真正的0m6.578s 用户0m0.015s sys 0m0.047s

按此速率，五个字节大约需要半个小时，六个字节应该占用一周的大部分时间。当然，计数器值对于那些人来说是不准确的 - 它主要用于验证较小字节值的迭代次数并给出循环要做的事情。

编辑：这是五个字节的时间，大概是我预测的半个小时：

$ time ./bytecounter a a a a a

使用5个字节，最终计数器值= 4294967295

真实的27m22.184s 用户0m0.015s sys 0m0.062s

Answer 6

好的，这里的代码是将任意十进制数作为第一个arg传递并从中倒数到零。我将其设置为允许计数器使用不同大小的元素（只需更改COUNTER_BASE的typedef），但事实证明，在我的系统上，字节实际上比短或长一些快。

#include <stdio.h>
#include <limits.h>  // defines CHAR_BIT
#include <ctype.h>
#include <vector>
using std::vector;
typedef unsigned char COUNTER_BASE;
typedef vector<COUNTER_BASE> COUNTER;
typedef vector<unsigned char> BYTEVEC;
const unsigned long byteMask = (~0ul) << CHAR_BIT;
const size_t MAXBYTES=20;

void mult10(BYTEVEC &val) {
    // Multiply value by 10
    unsigned int carry = 0;
    int i;
    for (i = 0; i < val.size(); i++) {
        unsigned long value = val[i]*10ul+carry;
        carry = (value & byteMask) >> CHAR_BIT;
        val[i] = value & ~byteMask;
    }
    if (carry > 0) val.push_back(carry);
}

void addDigit(BYTEVEC &val, const char digit) {
    // Add digit to the number in BYTEVEC.
    unsigned int carry = digit - '0'; // Assumes ASCII char set
    int i;
    for (i = 0; i < val.size() && carry; i++) {
        unsigned long value = static_cast<unsigned long>(val[i])+carry;
        carry = (value & byteMask) >> CHAR_BIT;
        val[i] = value & ~byteMask;
    }
    if (carry > 0) val.push_back(carry);
}

BYTEVEC Cstr2Bytevec(const char *str) {
    // Turn a C-style string into a BYTEVEC.  Only the digits in str apply,
    // so that one can use commas, underscores, or other non-digits to separate
    // digit groups.
    BYTEVEC result;
    result.reserve(MAXBYTES);
    result[0]=0;
    unsigned char *res=&result[0]; // For debugging
    while (*str) {
        if (isdigit(static_cast<int>(*str))) {
            mult10(result);
            addDigit(result, *str);
        }
        str++;
    }
    return result;
}

void packCounter(COUNTER &ctr, const BYTEVEC &val) {
    // Pack the bytes from val into the (possibly larger) datatype of COUNTER
    int i;
    ctr.erase(ctr.begin(), ctr.end());
    COUNTER_BASE value = 0;
    for (i = 0; i < val.size(); i++) {
        int pos = i%sizeof(COUNTER_BASE); // position of this byte in the value
        if (i > 0 && pos == 0) {
            ctr.push_back(value);
            value = val[i];
        } else {
            value |= static_cast<COUNTER_BASE>(val[i]) << pos*CHAR_BIT;
        }
    }
    ctr.push_back(value);
}

inline bool decrementAndTest(COUNTER &ctr) {
    // decrement value in ctr and return true if old value was not all zeros
    int i;
    for (i = 0; i < ctr.size() && !(ctr[i]--); i++) ; // EMPTY BODY
    return i < ctr.size();
}

inline bool decrementAndTest2(COUNTER_BASE *ctr, const size_t size) {
    // decrement value in ctr and return true if old value was not all zeros
    int i;
    for (i = 0; i < size && !(ctr[i]--); i++) ; // EMPTY BODY
    return i < size;
}

/* Vector counter - uses first arg (if supplied) as the count */
int main(int argc, const char *argv[]) {
    BYTEVEC limit = Cstr2Bytevec(argc > 1? argv[1] : "0");
    COUNTER ctr;
    packCounter(ctr, limit);
    COUNTER_BASE *ctr_vals = ctr.size() > 0 ? &ctr[0] : NULL; 
    size_t ctr_size = ctr.size();

    unsigned long ul_counter = 0ul; /* to give loop something to do */
    while(decrementAndTest2(ctr_vals, ctr_size)) {
        ul_counter++;
    };
    printf("With %d bytes used, final ul_counter value = %lu\n", limit.size(), ul_counter);
    return 0;
}

使用示例：

$ time ./bigcounter 5

使用1个字节，最终ul_counter值= 5

真正的0m0.094s 用户0m0.031s sys 0m0.047s

$ time ./bigcounter 5,000

使用2个字节，最终ul_counter值= 5000

真实0m0.062s 用户0m0.015s sys 0m0.062s

$ time ./bigcounter 5,000,000

使用3个字节，最终ul_counter值= 5000000

真正的0m0.093s 用户0m0.015s sys 0m0.046s

$ time ./bigcounter 1,000,000,000

使用4个字节，最终ul_counter值= 1000000000

真正的0m2.688s 用户0m0.015s sys 0m0.015s

$ time ./bigcounter 2,000,000,000

使用4个字节，最终ul_counter值= 2000000000

真正的0m5.125s 用户0m0.015s sys 0m0.046s

$ time ./bigcounter 3,000,000,000

使用4个字节，最终ul_counter值= 3000000000

真正的0m7.485s 用户0m0.031s sys 0m0.047s

$ time ./bigcounter 4,000,000,000

使用4个字节，最终ul_counter值= 4000000000

真正的0m9.875s 用户0m0.015s sys 0m0.046s

$ time ./bigcounter 5,000,000,000

使用5个字节，最终ul_counter值= 705032704

真正的0m12.594s 用户0m0.046s sys 0m0.015s

$ time ./bigcounter 6,000,000,000

使用5个字节，最终ul_counter值= 1705032704

真正的0m14.813s 用户0m0.015s sys 0m0.062s

将计数器向量展开到C风格的数据结构中（即使用decrementAndTest2而不是decrementAndTest）加速了大约20-25％，但代码仍然是我之前的类似大小的C程序的两倍慢例子（约40亿）。这是MS Visual C ++ 6.0作为发布模式下的编译器，针对两个程序在2GHz双核系统上优化速度。内联decrementAndTest2函数肯定会产生很大的差异（大约12秒，而50亿循环则为30），但是我必须看看是否在C程序中实际内联代码可以获得类似的性能。

Answer 7

主要功能中的变量甚至可以存储100个因子

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#include <map>
#include <functional>
#include <algorithm>
#include <cstdlib>
#include <iomanip>
#include <stack>
#include <queue>
#include <deque>
#include <limits>
#include <cmath>
#include <numeric>
#include <set>

using namespace std;


//template for BIGINIT

// base and base_digits must be consistent
const int base = 10;
const int base_digits = 1;

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);
            //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(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;
    }
};
 //use :  bigint var;
//template for biginit over





int main()
{  
   bigint var=10909000890789;
   cout<<var;
return 0;
}

在变量和循环中存储大数字

7 个答案: