题目链接：7-15 计算圆周率

一. 题目

1. 题目

2. 输入输出样例

3. 限制

二、代码

1. 代码实现

#include <stdio.h>

// 分子：阶乘
static unsigned long long int JieCheng (unsigned int n) {
    if (n == 1) {
        return 1;
    } else {
        return n * JieCheng(n - 1);
    }
}

// 分母：等差数列乘积，(2*0+1)*...* (2*n+1)
static unsigned long long int TwoNAdd1 (unsigned int n) {
    if (n == 1) {
        return 3;
    } else {
        return (2 * n + 1) * TwoNAdd1(n - 1);
    }
}

static float lastValue (unsigned int n) {
    return JieCheng(n) * 1.0 / TwoNAdd1(n);
}

int main(void) {
    unsigned int n;
    float limit, temp, result;

    if (scanf("%f", &limit) != 1) {
        return -1;
    }
    n = 1;
    temp = 1;
    result = 2;
    while (temp > limit) {
        temp = lastValue(n++);
        result += temp * 2;
    }
    printf("%.6f\n", result);
    return 0;
}

2. 提交结果

三、代码优化

1. 代码实现

#include <stdio.h>

// 分子：阶乘
static unsigned long long int JieCheng (unsigned int n) {
    unsigned int chengShu;
    if (n % 2) {
        chengShu = 1;
    } else {
        chengShu = n;
    }
    if (n == 1) {
        return 1;
    } else {
        return chengShu * JieCheng(n - 1);
    }
}

unsigned int N;

// 分母：等差数列乘积，(2*0+1)*...* (2*n+1)
static unsigned long long int TwoNAdd1 (unsigned int n) {
    unsigned int chengShu;
    chengShu = 2 * n + 1;
    if (chengShu <= N) { // 小于N的项，均存在分子中，约掉
        return 1;
    }
    if (n == 1) {
        return chengShu;
    } else {
        return chengShu * TwoNAdd1(n - 1);
    }
}

static float lastValue (unsigned int n) {
    return JieCheng(n) * 1.0 / TwoNAdd1(n);
}

int main(void) {
    float limit, temp, result;

    if (scanf("%f", &limit) != 1) {
        return -1;
    }
    N = 1;
    temp = 1;
    result = 2;
    while (temp > limit) { // 精度判断
        temp = lastValue(N);
        N++;
        result += temp * 2;
    }
    printf("%.6f\n", result);
    return 0;
}

2. 提交结果