【实践课外】6.循环结构3

【实践课外】6.循环结构3/7-1 二分法求多项式单根.c

@@ -0,0 +1,28 @@
+#include <stdio.h>
+
+int main() {
+    double a3, a2, a1, a0, a, b, ans = 0;
+
+    scanf("%lf%lf%lf%lf%lf%lf", &a3, &a2, &a1, &a0, &a, &b);
+
+    while (b - a > 1e-4) {
+        double mid = (a + b) / 2;
+        double f_a = a3 * a * a * a + a2 * a * a + a1 * a + a0;
+        double f_mid = a3 * mid * mid * mid + a2 * mid * mid + a1 * mid + a0;
+        double f_b = a3 * b * b * b + a2 * b * b + a1 * b + a0;
+
+        ans = mid;
+
+        if (-1e-4 < f_mid && f_mid < 1e-4) {  // 近似地认为 f(mid) == 0
+            break;
+        } else if (f_a * f_mid > 0) {  // 在 (mid, b) 中
+            a = mid;
+        } else {  // 在 (a, mid) 中
+            b = mid;
+        }
+    }
+
+    printf("%.2lf\n", ans);
+
+    return 0;
+}

【实践课外】6.循环结构3/7-2 循环-十进制转化.c

@@ -0,0 +1,19 @@
+#include <stdio.h>
+
+int main() {
+    int n, m;
+
+    scanf("%d%d", &n, &m);
+
+    int result = 0, factor = 1;
+
+    while (n) {
+        result += (n % m) * factor;
+        n /= m;
+        factor *= 10;
+    }
+
+    printf("%d\n", result);
+
+    return 0;
+}

【实践课外】6.循环结构3/7-3 梅森数.c

@@ -0,0 +1,32 @@
+#include <stdio.h>
+
+int main() {
+    int n, num = 2, count = 0;
+
+    scanf("%d", &n);
+
+    for (int i = 2; i <= n; i++) {
+        num *= 2;
+
+        int is_prime = 1;
+
+        for (int j = 2; j * j <= num - 1; j++) {
+            if ((num - 1) % j == 0) {
+                is_prime = 0;
+
+                break;
+            }
+        }
+
+        if (is_prime == 1) {
+            printf("%d\n", num - 1);
+
+            count++;
+        }
+    }
+    if (count == 0) {
+        printf("None\n");
+    }
+
+    return 0;
+}

【实践课外】6.循环结构3/7-4 单词长度.c

@@ -0,0 +1,39 @@
+#include <stdio.h>
+
+int main() {
+    char c;
+    int count = 0;
+    int printed = 0;
+
+    while (scanf("%c", &c) != EOF, c != '.') {
+        if (c == ' ') {
+            if (count != 0) {
+                if (printed == 0) {
+                    printf("%d", count);
+
+                    printed = 1;
+                } else {
+                    printf(" %d", count);
+                }
+            }
+
+            count = 0;
+        } else {
+            count++;
+        }
+    }
+
+    if (count != 0) {
+        if (printed == 0) {
+            printf("%d", count);
+
+            printed = 1;
+        } else {
+            printf(" %d", count);
+        }
+    }
+
+    printf("\n");
+
+    return 0;
+}

【实践课外】6.循环结构3/7-5 21循环-求和3.c

@@ -0,0 +1,21 @@
+#include <stdio.h>
+
+int main() {
+    int n, k, ans = 0;
+
+    scanf("%d%d", &n, &k);
+
+    for (int i = 1; i <= n; i++) {
+        int tmp = 1;
+
+        for (int j = 1; j <= k; j++) {
+            tmp = tmp * i % 114514;
+        }
+
+        ans = (ans + tmp) % 114514;
+    }
+
+    printf("%d\n", ans);
+
+    return 0;
+}

【实践课外】6.循环结构3/7-6 21循环-金字塔.c

@@ -0,0 +1,39 @@
+#include <stdio.h>
+
+int main() {
+    int n, sum = 0, max_level;
+
+    scanf("%d", &n);
+
+    // 计算可以摆出的最大金字塔层数
+    for (max_level = 1; sum + max_level <= n; max_level++) {
+        sum += max_level;
+    }
+
+    max_level--;  // 循环结束时 i 多加了一次
+
+    // 自顶向下
+    for (int j = max_level; j >= 1; j--) {
+        // 左侧的 0
+        for (int k = j; k > 1; k--) {
+            printf("0");
+        }
+
+        // 当前层的数字
+        printf("%d", j);
+
+        // 中间的0和当前层的数字
+        for (int m = j; m < max_level; m++) {
+            printf("0%d", j);
+        }
+
+        // 右侧的 0
+        for (int k = j; k > 1; k--) {
+            printf("0");
+        }
+
+        printf("\n");
+    }
+
+    return 0;
+}

【实践课外】6.循环结构3/7-7 循环-杨辉三角.c

@@ -0,0 +1,41 @@
+#include <stdio.h>
+
+int n, m, a[25][25];
+
+int main() {
+    scanf("%d%d", &n, &m);
+
+    if (n > m) {
+        int t = n;
+        n = m;
+        m = t;
+    }
+
+    // 递推法计算杨辉三角
+    for (int i = 1; i <= m; i++) {
+        a[i][1] = a[i][i] = 1;
+
+        for (int j = 2; j < i; j++) {
+            a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
+        }
+    }
+
+    for (int i = n; i <= m; i++) {
+        // 输出空格
+        for (int j = 1; j <= m - i; j++) {
+            printf(" ");
+        }
+
+        for (int j = 1; j <= i; j++) {
+            printf("%d", a[i][j] % 10);
+
+            if (j != i) {
+                printf(" ");
+            }
+        }
+
+        printf("\n");
+    }
+
+    return 0;
+}