C_TA_Website/source/_posts/answer4.md

title	mathjax	date	tags
第四次作业（答案）	true	2020-11-06 09:57:17

P137-3

#include <stdio.h>
#include <stdlib.h>

int main()
{
	// 输入
	printf("最大公约数与最小公倍数计算");
	int m, n;
	printf("请输入正整数m和n：\nm = ");
	if (scanf("%d", &m) != 1 || m <= 0) {
		fprintf(stderr, "非法输入！\n");
		exit(EXIT_FAILURE);
	}
	printf("n = ");
	if (scanf("%d", &n) != 1 || n <= 0) {
		fprintf(stderr, "非法输入！\n");
		exit(EXIT_FAILURE);
	}

	// a为较大者，b为较小者
	int a, b;
	if (m > n) {
		a = m;
		b = n;
	} else {
		a = n;
		b = m;
	}

	// 求最大公因数
	int t = 1;
	while (t != 0) {
		t = a % b;
		a = b;
		b = t;
	}

	// 最小公倍数
	int lcm = m * n / a;

	printf("最大公因数：%d\n最小公倍数：%d\n", a, lcm);
    return 0;
}

P137-5

\begin{align*} S_{n} &= a + aa + aaa + \cdots + \overbrace{aa \cdots a}^{n个a} \ &= a \times 10^{n - 1} + 2a \times 10^{n - 2} + \cdots + na \ &= (((a \times 10 + 2a) \times 10 + 3a) \times 10 + \cdots ) + na \end{align*}

由此，我们编写程序：

#include <stdio.h>

int main()
{
	int a, n;
	printf("P137-5\na = ");
	scanf("%d", &a);
	printf("n = ");
	scanf("%d", &n);

	int sum = a;

	for (int i = 2; i <= n; ++i) {
		sum *= 10;
		sum += a * i;
	}
	printf("S = %d\n", sum);
	return 0;
}

P138-14

迭代公式可表示为：

\begin{equation*} x_{n + 1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \end{equation*}

设

\begin{equation*} f(x) = 2x^{3} - 4x^{2} + 3 x - 6 \end{equation*}

则

\begin{align*} f'(x) &= 6x^{2} - 8x + 3 \ x_{n + 1} &= x_{n} - \frac{f(x_{n})}{f'(x_{n})} \ &= x_{n} - \frac{2x_{n}^{3} - 4x_{n}^{2} + 3x_{n} - 6}{6x_{n}^{2} - 8x_{n} + 3} \ &= \frac{4x_{n}^{3} - 4x_{n}^{2} + 6}{6x_{n}^{2} - 8x_{n} + 3} \ \end{align*}

不妨设迭代的终止条件为：

\begin{equation*} |f(x_{n})| < 10^{-6} \end{equation*}

此时求得的 x_{n} 可以作为一个近似解。

#include <stdio.h>
#include <math.h>

int main()
{
	double t = 1e-6;
	double x = 1.5;
	while(fabs(2 * pow(x, 3) - 4 * pow(x, 2) + 3 * x - 6) >= t) {
		x = (4 * pow(x, 3) - 4 * pow(x, 2) + 6) / (6 * pow(x, 2) - 8 * x + 3);
	}
	printf("P138-14\nx = %.3f\n", x);
	return 0;
}

P138-15

#include <stdio.h>
#include <math.h>

int main()
{
	double a = -10, b = 10, c = 0;
	double t = 1e-6;
	double fa = 2 * pow(a, 3) - 4 * pow(a, 2) + 3 * a - 6;
	double fc = 2 * pow(c, 3) - 4 * pow(c, 2) + 3 * c - 6;
	while (fabs(2 * pow(c, 3) - 4 * pow(c, 2) + 3 * c - 6) >= t) {
		if (fabs(fc) < t) {
			break;
		}
		if (fa * fc < 0) {
			b = c;
		} else {
			a = c;
			fa = fc;
		}
		c = (a + b) / 2;
		fc = 2 * pow(c, 3) - 4 * pow(c, 2) + 3 * c - 6;
	}
	printf("P138-15\nx = %.3f\n", c);
	return 0;
}

选择排序

#include <stdio.h>
#include <math.h>

#define LEN 10

int main()
{
	printf("请输入%d个数字，用空格分隔：\n", LEN);
	int a[LEN];
	for (int i = 0; i < LEN; ++i) {
		scanf("%d", a + i);
	}
	for (int i = 0; i < LEN - 1; ++i) {
		int m = i;
		for (int j = i + 1; j < LEN; ++j) {
			if (a[j] < a[m]) {
				m = j;
			}
		}
		if (m != i) {
			int tmp = a[i];
			a[i] = a[m];
			a[m] = tmp;
		}
	}
	for (int i = 0; i < LEN; ++i) {
		printf("%d ", a[i]);
	}
	printf("\n");
	return 0;
}