上一次我们说到了这个程序

#include <iostream>  
#include <cmath>  
#include <limits>  
  

int continuedFractionTerm(int n) {  
    if (n == 0) return 1;  
    if (n % 2 == 0) {  
        return 2 * n + 1;  
    } else {  
        return 2 * n;  
    }  
}  
  

std::pair<int, int> bestRationalApproximation(double target, int maxDenominator) {  
    int numerator = 0;  
    int denominator = 1;  
    int prevNumerator = 1;  
    int prevDenominator = 0;  
    double bestDiff = std::numeric_limits<double>::max();  
  
    for (int i = 0; i < maxDenominator; ++i) {  
        int term = continuedFractionTerm(i);  
        int newNumerator = prevDenominator + term * prevNumerator;  
        int newDenominator = prevNumerator;  
        prevNumerator = newNumerator;  
        prevDenominator = newDenominator;  
  
        double currentPi = static_cast<double>(newNumerator) / newDenominator;  
        double diff = std::abs(currentPi - target);  
  
        if (diff < bestDiff) {  
            bestDiff = diff;  
            numerator = newNumerator;  
            denominator = newDenominator;  
        }  
    }  
  
    return {numerator, denominator};  
}  
  
int main() {  
    const double pi_approx = 3.141592653897932384;  
    int maxDenominator = 100000000; 
  
    std::pair<int, int> approximation = bestRationalApproximation(pi_approx, maxDenominator);  
  
    std::cout << "最接近π的分数是: " << approximation.first << "/" << approximation.second << std::endl;  
    std::cout << "差异是: " << std::abs(static_cast<double>(approximation.first) / approximation.second - pi_approx) << std::endl;  
  
    return 0;  
}

这个是比较快的，还有一个我没测过（准确的说我是没有那个耐心等）

(￣_,￣ ) 

#include <iostream>  
#include <cmath>  
#include <limits>  
  
int main() {  
    const double pi_approx = 3.141592653589793; // π的近似值  
    int numerator = 0; // 分子  
    int denominator = 1; // 分母  
    double best_diff = std::numeric_limits<double>::max(); // 初始差异设置为最大double值  
  
    // 迭代地检查分数  
    for (int i = 1; i < 1000000; ++i) { // 限制迭代次数以防止无限循环  
        for (int j = 1; j < 1000000; ++j) {  
            double current_pi = static_cast<double>(i) / j; // 当前分数表示的π值  
            double diff = std::abs(current_pi - pi_approx); // 计算差异  
  
            // 如果当前分数更接近π，则更新最佳分数和差异  
            if (diff < best_diff) {  
                best_diff = diff;  
                numerator = i;  
                denominator = j;  
            }  
        }  
    }  
  
    std::cout << "最接近π的分数是: " << numerator << "/" << denominator << std::endl;  
    std::cout << "差异是: " << best_diff << std::endl;  
  
    return 0;  
}

 长度确实短了一点，但是时间真长。

完！