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在学习 大数n!算法时, 想到log2(x)怎么算?

math.h中没有log2接口, 只有log和log10.

上网查了下, 原来用对数换底公式可以搞定这事^_^

// @file power_exp_log.cpp// @brief 计算以2为底的log#include 
   
                     // for i/o functions#include 
    
                        // for exp(), log(), and log10()#include 
     
      using namespace std;#define CONST_E 2.718282 ///< 常数e的值约为2.718282/// math.h中没有log2的函数double log2(double dblpow2);void fnTest_power_exp_log_2(); ///< 用对数换底公式做void fnTest_power_exp_log_E();void fnTest_power_exp_log_10();int main(int argc, char* argv[]){    fnTest_power_exp_log_2();    // fnTest_power_exp_log_E();    // fnTest_power_exp_log_10();    printf("END, press any key to quit\n");    _getch();    return 0;}double log2(double dblpow2){    /**    math.h中没有log2的接口    用对数换底公式来间接计算    对数换底公式    Loga(B) = logc(B) / logc(A)    a, c均大于0且不等于1    math.h中有log和log10    logc 可以用log或1og10    */    double dbllog2 = 0;    dbllog2 = log(dblpow2) / log(2);    return dbllog2;}void fnTest_power_exp_log_2(){    double dblIndex = 0;    double dblpow = 0;    double dbllog = 0;    for (dblIndex = 0; dblIndex <= 10; dblIndex++)    {        dblpow = pow(2, dblIndex);        printf("pow(2, %f) = %f\n",             dblIndex, dblpow);        dbllog = log2(dblpow);        printf("log2(%f) = %f\n",             dblpow, dbllog);        printf("\r\n");    }    /** run result    pow(2, 0.000000) = 1.000000    log2(1.000000) = 0.000000    pow(2, 1.000000) = 2.000000    log2(2.000000) = 1.000000    pow(2, 2.000000) = 4.000000    log2(4.000000) = 2.000000    pow(2, 3.000000) = 8.000000    log2(8.000000) = 3.000000    pow(2, 4.000000) = 16.000000    log2(16.000000) = 4.000000    pow(2, 5.000000) = 32.000000    log2(32.000000) = 5.000000    pow(2, 6.000000) = 64.000000    log2(64.000000) = 6.000000    pow(2, 7.000000) = 128.000000    log2(128.000000) = 7.000000    pow(2, 8.000000) = 256.000000    log2(256.000000) = 8.000000    pow(2, 9.000000) = 512.000000    log2(512.000000) = 9.000000    pow(2, 10.000000) = 1024.000000    log2(1024.000000) = 10.000000    */}void fnTest_power_exp_log_10(){    double dblIndex = 0;    double dblpow = 0;    double dbllog = 0;    for (dblIndex = 0; dblIndex <= 10; dblIndex++)    {        dblpow = pow(10, dblIndex);        printf("pow(10, %f) = %f\n",             dblIndex, dblpow);        dbllog = log10(dblpow);        printf("log10(%f) = %f\n",             dblpow, dbllog);        printf("\r\n");    }    /** run result    pow(10, 0.000000) = 1.000000    log10(1.000000) = 0.000000    pow(10, 1.000000) = 10.000000    log10(10.000000) = 1.000000    pow(10, 2.000000) = 100.000000    log10(100.000000) = 2.000000    pow(10, 3.000000) = 1000.000000    log10(1000.000000) = 3.000000    pow(10, 4.000000) = 10000.000000    log10(10000.000000) = 4.000000    pow(10, 5.000000) = 100000.000000    log10(100000.000000) = 5.000000    pow(10, 6.000000) = 1000000.000000    log10(1000000.000000) = 6.000000    pow(10, 7.000000) = 10000000.000000    log10(10000000.000000) = 7.000000    pow(10, 8.000000) = 100000000.000000    log10(100000000.000000) = 8.000000    pow(10, 9.000000) = 1000000000.000000    log10(1000000000.000000) = 9.000000    pow(10, 10.000000) = 10000000000.000000    log10(10000000000.000000) = 10.000000    */}void fnTest_power_exp_log_E(){    double dblIndex = 0;    double dblexp = 0;    double dblpow = 0;    double dbllog = 0;    for (dblIndex = 0; dblIndex <= 10; dblIndex++)    {        dblpow = pow(CONST_E, dblIndex);        printf("pow(CONST_E, %f) = %f\n",             dblIndex, dblpow);        dblexp = exp(dblIndex);        printf("exp(%f) = %f\n",             dblIndex, dblexp);        dbllog = log(dblexp);        printf("log(%f) = %f\n",             dblexp, dbllog);        printf("\r\n");    }    /** run result    pow(CONST_E, 0.000000) = 1.000000    exp(0.000000) = 1.000000    log(1.000000) = 0.000000    pow(CONST_E, 1.000000) = 2.718282    exp(1.000000) = 2.718282    log(2.718282) = 1.000000    pow(CONST_E, 2.000000) = 7.389057    exp(2.000000) = 7.389056    log(7.389056) = 2.000000    pow(CONST_E, 3.000000) = 20.085541    exp(3.000000) = 20.085537    log(20.085537) = 3.000000    pow(CONST_E, 4.000000) = 54.598164    exp(4.000000) = 54.598150    log(54.598150) = 4.000000    pow(CONST_E, 5.000000) = 148.413206    exp(5.000000) = 148.413159    log(148.413159) = 5.000000    pow(CONST_E, 6.000000) = 403.428946    exp(6.000000) = 403.428793    log(403.428793) = 6.000000    pow(CONST_E, 7.000000) = 1096.633643    exp(7.000000) = 1096.633158    log(1096.633158) = 7.000000    pow(CONST_E, 8.000000) = 2980.959492    exp(8.000000) = 2980.957987    log(2980.957987) = 8.000000    pow(CONST_E, 9.000000) = 8103.088530    exp(9.000000) = 8103.083928    log(8103.083928) = 9.000000    pow(CONST_E, 10.000000) = 22026.479695    exp(10.000000) = 22026.465795    log(22026.465795) = 10.000000    */}
     
    
   

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