From 56c07da12c0af31cac7fd9588bd8b905807e548c Mon Sep 17 00:00:00 2001
From: Zhao Zuohong <1040110848@qq.com>
Date: Fri, 6 Nov 2020 09:57:35 +0800
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+---
+title: 第四次作业（答案）
+mathjax: true
+date: 2020-11-06 09:57:17
+tags:
+---
+
+
+# P137-3
+
+```C
+#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：
+
+```C
+#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。
+
+```C
+#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
+
+```C
+#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
+
+```C
+#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;
+}
+```