如何實現快速將 64 位二進位(存在字元串里)轉換為十進位？

01-02

不使用long或long long

更新：之前缺少了進位處理。

我理解這個題目是說，不轉換成64位整數，直接把二進位的字元串轉換成十進位。

設二進位為 $b_{63}b_{62}ldots b_1b_0$ ，那麼第 j 個十進位為

$d_j = left lfloor frac{sum 2^i b_i}{10^j} ight floor mod 10$

為了直接計算二進位各位的和，可以轉換成以下的形式（ $c_j$ 為進位）

$d_j + c_j= sum b_i left (left lfloor frac{ 2^i }{10^j} ight floor mod 10 ight)$

那麼，我們只需要預計算中間那個基於(i, j) 的表（每個元素的值 &<10 ），就可以求出每個十進位的值：

#include & #include & #include & #include &


static uint8_t weight[64 * 19] = {

1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

8,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

2,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

4,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

8,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

6,5,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

2,1,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

4,2,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

8,4,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

6,9,0,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

2,9,1,8,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

4,8,3,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

8,6,7,2,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

6,3,5,5,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

2,7,0,1,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,

4,4,1,2,6,2,0,0,0,0,0,0,0,0,0,0,0,0,0,

8,8,2,4,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,

6,7,5,8,4,0,1,0,0,0,0,0,0,0,0,0,0,0,0,

2,5,1,7,9,0,2,0,0,0,0,0,0,0,0,0,0,0,0,

4,0,3,4,9,1,4,0,0,0,0,0,0,0,0,0,0,0,0,

8,0,6,8,8,3,8,0,0,0,0,0,0,0,0,0,0,0,0,

6,1,2,7,7,7,6,1,0,0,0,0,0,0,0,0,0,0,0,

2,3,4,4,5,5,3,3,0,0,0,0,0,0,0,0,0,0,0,

4,6,8,8,0,1,7,6,0,0,0,0,0,0,0,0,0,0,0,

8,2,7,7,1,2,4,3,1,0,0,0,0,0,0,0,0,0,0,

6,5,4,5,3,4,8,6,2,0,0,0,0,0,0,0,0,0,0,

2,1,9,0,7,8,6,3,5,0,0,0,0,0,0,0,0,0,0,

4,2,8,1,4,7,3,7,0,1,0,0,0,0,0,0,0,0,0,

8,4,6,3,8,4,7,4,1,2,0,0,0,0,0,0,0,0,0,

6,9,2,7,6,9,4,9,2,4,0,0,0,0,0,0,0,0,0,

2,9,5,4,3,9,9,8,5,8,0,0,0,0,0,0,0,0,0,

4,8,1,9,6,8,9,7,1,7,1,0,0,0,0,0,0,0,0,

8,6,3,8,3,7,9,5,3,4,3,0,0,0,0,0,0,0,0,

6,3,7,6,7,4,9,1,7,8,6,0,0,0,0,0,0,0,0,

2,7,4,3,5,9,8,3,4,7,3,1,0,0,0,0,0,0,0,

4,4,9,6,0,9,7,7,8,4,7,2,0,0,0,0,0,0,0,

8,8,8,3,1,8,5,5,7,9,4,5,0,0,0,0,0,0,0,

6,7,7,7,2,6,1,1,5,9,9,0,1,0,0,0,0,0,0,

2,5,5,5,5,2,3,2,0,9,9,1,2,0,0,0,0,0,0,

4,0,1,1,1,5,6,4,0,8,9,3,4,0,0,0,0,0,0,

8,0,2,2,2,0,3,9,0,6,9,7,8,0,0,0,0,0,0,

6,1,4,4,4,0,6,8,1,2,9,5,7,1,0,0,0,0,0,

2,3,8,8,8,0,2,7,3,4,8,1,5,3,0,0,0,0,0,

4,6,6,7,7,1,4,4,7,8,6,3,0,7,0,0,0,0,0,

8,2,3,5,5,3,8,8,4,7,3,7,0,4,1,0,0,0,0,

6,5,6,0,1,7,6,7,9,4,7,4,1,8,2,0,0,0,0,

2,1,3,1,2,4,3,5,9,9,4,9,2,6,5,0,0,0,0,

4,2,6,2,4,8,6,0,9,9,9,8,5,2,1,1,0,0,0,

8,4,2,5,8,6,3,1,8,9,9,7,1,5,2,2,0,0,0,

6,9,4,0,7,3,7,2,6,9,9,5,3,0,5,4,0,0,0,

2,9,9,0,4,7,4,5,2,9,9,1,7,0,0,9,0,0,0,

4,8,9,1,8,4,9,0,5,8,9,3,4,1,0,8,1,0,0,

8,6,9,3,6,9,8,1,0,7,9,7,8,2,0,6,3,0,0,

6,3,9,7,2,9,7,3,0,4,9,5,7,5,0,2,7,0,0,

2,7,8,5,5,8,5,7,0,8,8,1,5,1,1,4,4,1,0,

4,4,7,1,1,7,1,5,1,6,7,3,0,3,2,8,8,2,0,

8,8,4,3,2,4,3,0,3,2,5,7,0,6,4,6,7,5,0,

6,7,9,6,4,8,6,0,6,4,0,5,1,2,9,2,5,1,1,

2,5,9,3,9,6,3,1,2,9,0,0,3,4,8,5,0,3,2,

4,0,9,7,8,3,7,2,4,8,1,0,6,8,6,1,1,6,4,

8,0,8,5,7,7,4,5,8,6,3,0,2,7,3,3,2,2,9,

};
// s must be [01]{1,64}

std::string bin2dec(const std::string s) {

    assert(s.length() &>= 0  s.length() &<= 64);
    uint8_t sum[20] = { 0 };
    for (int i = 0; i &< s.length(); i++)
        if (s[i] == "1")
            for (int j = 0; j &< 19; j++) {
                sum[j] += weight[(s.length() - 1 - i) * 19 + j];
                for (int c = j; sum[c] &>= 10; c++) {

                    sum[c] -= 10;

                    sum[c + 1]++;

                }

            }

        else

            assert(s[i] == "0");
    // First non-zero digit

    int k = 19;

    while (k &> 0  sum[k] == 0)

        k--;

// reverse output std::string o; o.reserve(k + 1); for (int j = 0; j &<= k; j++) o.push_back(sum[k - j] + "0"); return o; } int main(int argc, char* argv[]) { // Generate weight table // uint64_t divisor = 1; // for (int j = 0; j &< 19; j++) { // for (int i = 0; i &< 64; i++) { // weight[i * 19 + j] = static_cast&(((uint64_t(1) &<&< i) / divisor) % 10); // } // divisor *= 10; // } // for (int i = 0; i &< 64; i++) { // for (int j = 0; j &< 19; j++) // std::cout &<&< (int)weight[i * 19 + j] &<&< ","; // std::cout &<&< std::endl; // } std::cout &<&< bin2dec(argv[1]) &<&< std::endl; }

bin2dec() 中避免了使用除法和模除，它使用減法來實現算式中第二個模除。

可以看到，表裡有很多 0 ，其實可以再做一個表去表示真正需要的長度，大家可以嘗試一下。

另外，這個做法實際上可能比正常做法（先轉成整數再轉成10進數字元串）為慢，但可以免去除法和模除，或適合超過64位而又有限長度的輸入。

#include &

// std::string binary // unsigned long long value value = std::stoull(binary, nullptr, 2);

或者

#include &

// std::string binary // unsigned long long value value = std::stroull(binary.c_str(), nullptr, 2);

兩者沒有大區別，卻需要有不同的出錯處理。

# 抱歉，下面答成十六進位轉換了。請忽略。

C++ 風格

#include & #include &

// std::string hex // uint64_t value std::istringstream(hex) &>&> std::hex &>&> value;

C 風格

#include & #include &

// std::string hex // uint64_t value std::sscanf(hex.c_str(), "%" SCNx64, value);

#include & #include & #include &


unsigned long convert_to_ulong(const std::string str)

{

    return std::bitset&(str).to_ulong();

}