位阵列除法的其余部分 C++

下面是 std：：bitset 的实现。

找到余数涉及将除数向左移动，直到它大于或等于除数，然后开始将其移回右侧，直到它处于原始位置。对于移位除数的每个新值，如果它大于运行余数（以除数开始），则减去它以获得新的余数。 如果除数永远等于余数，则返回 0。一旦除数到达其原始的、未移动的位置，如果需要，从余数中减去后，余数就完成了。

BsMod函数采用除数和除数参数，除数参数替换为余数，因此请确保该参数是左值，以便获得结果。

默认测试（在 main 中）随机创建二进制字符串并打印出二进制结果。 这有点难以验证，所以我做了另一个随机测试（Test()），它使用整数值来自动验证结果。

#include <iostream>
#include <bitset>
#include <string>
#include <ctime>
#include <cstdlib>
// subtracts b from a, replacing a with the result
template <typename A, typename B>
void Subtract(A &a, const B &b) {
    static const std::size_t minc = a.size() < b.size() ? a.size() : b.size();
    bool borrow = false;
    for(std::size_t i = 0; i<minc; ++i) {
        const bool dif = a[i] ^ b[i] ^ borrow;
        borrow = (a[i] && b[i] && borrow) || (!a[i] && (b[i] || borrow));
        a[i] = dif;
    }
    for(std::size_t i=minc; borrow && i<a.size(); ++i) {
        a[i] = borrow = !a[i];
    }
}
// Returns the index of the highest set bit in b
// Returns unsigned -1 if all bits are 0
template <typename B>
std::size_t HiBit(const B& b) {
    for(std::size_t i = b.size()-1; i+1; --i) {
        if(b[i]) return i;
    }
    // b is zero
    return ~std::size_t(0);
}
// Compare returns 1 if a>b, 0 if a==b or -1 if a<b
template <typename B>
int Compare(const B &a, const B &b) {
    const std::size_t high = a.size()-1;
    for(std::size_t i=high; i+1; --i) {
        if(a[i] != b[i]) {
            return int(a[i]) - int(b[i]);
        }
    }
    return 0;
}
// nr is changed from the dividend to the remainder
template <typename B>
void BsMod(B &nr, B d) {
    const std::size_t hi_n = HiBit(nr);
    const std::size_t hi_d = HiBit(d);
    if(hi_d > hi_n) return;                            // nr < d, keep n as r
    if(hi_d == hi_n && Compare(nr, d) == -1) return;   // nr < d, keep n as r
    const std::size_t dshift = hi_n - hi_d;
    d <<= dshift;
    for(std::size_t i=0; i<=dshift; ++i) {
        const int cmp = Compare(nr, d);
        if(cmp == 0) { nr = B(); return; } // d evenly divides nr, so r is 0
        if(cmp > 0) {  // nr > shifted d
            // the quotient would accumulate a 1 bit here, at the d shift position
            Subtract(nr, d);
        }
        d >>= 1;   // divide d by 2, shift back toward original position
    }
}
template <typename B>
unsigned long long bs_to_ull(const B& b) {
    unsigned long long result = 0;
    for(std::size_t i=0; i<sizeof(unsigned long long)*8; ++i) {
        result |= static_cast<unsigned long long>(b[i]) << i;
    }
    return result;
}
template <typename B>
void ull_to_bs(B& b, unsigned long long n) {
    b.reset();
    for(std::size_t i=0; i<sizeof(unsigned long long)*8; ++i) {
        if(n & ((unsigned long long)1 << i)) b.set(i, true);
    }
}
unsigned long long rand_ull() {
    unsigned long long r = 0;
    unsigned long long b = 0;
    for(int i=0; i<sizeof(unsigned long long); ++i) {
        r = r * 33 + rand();
        b ^= rand();
        b <<= 8;
    }
    return ((r << sizeof(unsigned long long)*4) | (r >> sizeof(unsigned long long )*4)) ^ b;
}
void Test(unsigned long long max=0, int max_iters=0) {
    typedef unsigned long long uit;
    typedef std::bitset<sizeof(unsigned long long)*8+8> bs;
    typedef unsigned long long uit;
    int iter_count = 0;
    for(;;) {
        uit a = rand_ull();
        uit b = rand_ull();
        if(max) {
            a %= max;
            b %= max;
        }
        if(rand() & 255) {
            while(b > a) b >>= rand() & 3;
        }
        if(!b) continue;
        bs bsa;
        ull_to_bs(bsa, a);
        bs bsb;
        ull_to_bs(bsb, b);
        BsMod(bsa, bsb);
        uit ibsm = bs_to_ull(bsa);
        uit m = a % b;
        std::cout << a << " % " << b << " = " << m << "    :    " << ibsm << 'n';
        if(ibsm != m) {
            std::cout << "Errorn";
            return;
        }
        ++iter_count;
        if(max_iters && iter_count > max_iters) break;
    }
}
std::string RandomBinaryString(unsigned bit_count) {
    std::string binstr;
    for(unsigned i=0; i<bit_count; ++i) {
        binstr +=  ((rand() >> (i%5)) ^ i) & 1 ?  '1' : '0';
    }
    return binstr;
}
void TrimLeadingZeros(std::string& s) {
    if(s.length() < 2 || s[0] != '0') return;
    for(std::string::size_type i=1; i<s.length()-1; ++i) {
        if(s[i] != '0') {
            s = s.substr(i);
            return;
        }
    }
    s = s.substr(s.length()-1);
}
int main() {
    srand((unsigned int)time(0));
    //Test(0, 10);  // test with integer values (which are easy to auto-validate)
    //return 0;
    std::string a = RandomBinaryString(200);
    std::string b = RandomBinaryString(10);
    static const int max_bitount = 220;
    typedef std::bitset<max_bitount> bs;
    bs bsa(a);
    bs bsb(b);
    // both arguments must have the same type (number of bits)
    // bsa gets replaced with bsa modulo bsb
    BsMod(bsa, bsb); 

    std::string c = bsa.to_string();
    TrimLeadingZeros(c);
    std::cout << a << "n   modn" << b << "n   ==n" << c << 'n';
}

我首先看一下使用 std：：bitset。这似乎比数组更容易使用。其次，阅读有关使用按位运算执行取模的文章。我找到的其中一篇文章是这样的。祝你好运。

首先为两种类型的数字（200 位和 10 位）定义移位减法。事后起诉该司。