1.编写程序,在键盘上输入一个n值,输出n以内所有个位-|||-数为3,但不能被3整除的数并求出它们的和。-|||-要求:-|||-输入:100-|||-输出:13,23,23,53,73,83, sum=288用C语言程序编写

### 打印 ASCII 码表 以下是 C 语言实现打印 ASCII 码表的代码示例： ```c #include <stdio.h> int main() { int i; for (i = 0; i <= 255; i++) { printf("ASCII value of %d is '%c'\n", i, i); } return 0; } ``` --- ### 输出素数 以下是一个优化后的素数输出程序，基于 `sqrt()` 函数[^3]： ```c #include <stdio.h> #include <math.h> int main() { int n = 100, i, j, flag; for (i = 2; i <= n; i++) { flag = 1; for (j = 2; j <= sqrt(i); j++) { if (i % j == 0) { flag = 0; break; } } if (flag == 1) { printf("%d ", i); } } return 0; } ``` --- ### 打印乘法表 根据用户需求动态调整行数的乘法表代码如下[^2]： ```c #include <stdio.h> void print_multiplication_table(int lines) { int i, j; for (i = 1; i <= lines; i++) { for (j = 1; j <= i; j++) { if (lines >= 10 && lines < 20) { printf("%d*%d=%-4d", j, i, i * j); } else if (lines >= 20) { printf("%d*%d=%-6d", j, i, i * j); } else { printf("%d*%d=%-2d", j, i, i * j); } } printf("\n"); } } int main() { int lines; printf("Enter the number of rows: "); scanf("%d", &lines); print_multiplication_table(lines); return 0; } ``` --- ### 数据统计 以下是对一组整数进行简单统计数据（最大值、最小值、平均值）的代码： ```c #include <stdio.h> int main() { int numbers[10], max, min, sum = 0, i; float avg; printf("Enter 10 integers:\n"); for (i = 0; i < 10; i++) { scanf("%d", &numbers[i]); sum += numbers[i]; } max = min = numbers[0]; for (i = 1; i < 10; i++) { if (max < numbers[i]) max = numbers[i]; if (min > numbers[i]) min = numbers[i]; } avg = (float)sum / 10; printf("Max: %d\nMin: %d\nAverage: %.2f\n", max, min, avg); return 0; } ``` --- ### 数字倒置 以下是从键盘输入一个正整数并将其反转的代码： ```c #include <stdio.h> int reverse_number(int num) { int reversed = 0; while (num != 0) { reversed = reversed * 10 + num % 10; num /= 10; } return reversed; } int main() { int num; printf("Enter a positive integer: "); scanf("%d", &num); printf("Reversed Number: %d\n", reverse_number(num)); return 0; } ``` --- ### 计算 e 的近似值 通过泰勒展开公式计算自然常数 \(e\) 的近似值[^4]： ```c #include <stdio.h> double calculate_e(int terms) { double sum = 0.0, temp = 1.0; int i; for (i = 1; i <= terms; i++) { temp *= i; sum += 1.0 / temp; } return sum; } int main() { int terms; printf("Enter the number of terms to approximate e: "); scanf("%d", &terms); printf("Approximation of e with %d terms: %.8lf\n", terms, calculate_e(terms)); return 0; } ``` --- ### cos(x) 近似值 使用泰勒级数计算 \(\cos(x)\) 的近似值： ```c #include <stdio.h> #include <math.h> double cosine_approx(double x, int terms) { double term = 1.0, result = 0.0; int sign = 1; for (int i = 0; i < terms; i++) { result += sign * term; sign = -sign; term *= (x * x) / ((2 * i + 2) * (2 * i + 3)); } return result; } int main() { double angle; int terms; printf("Enter angle in radians and number of terms: "); scanf("%lf%d", &angle, &terms); printf("Cosine approximation: %.8lf\n", cosine_approx(angle, terms)); return 0; } ``` --- ### 解方程分配问题 假设问题是线性方程组求解，以下是一个简单的高斯消元法实现： ```c #include <stdio.h> #define N 3 void gauss_jordan(float matrix[N][N + 1], float solution[N]) { int i, j, k; for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { if (i != j) { float factor = matrix[j][i] / matrix[i][i]; for (k = 0; k <= N; k++) { matrix[j][k] -= factor * matrix[i][k]; } } } } for (i = 0; i < N; i++) { solution[i] = matrix[i][N] / matrix[i][i]; } } int main() { float matrix[N][N + 1] = {{2, 1, -1, 8}, {-3, -1, 2, -11}, {-2, 1, 2, -3}}; float solutions[N]; gauss_jordan(matrix, solutions); printf("Solutions: x=%.2f, y=%.2f, z=%.2f\n", solutions[0], solutions[1], solutions[2]); return 0; } ``` --- ### 寻找完数 以下是一个寻找指定范围内的完全数的程序： ```c #include <stdio.h> void find_perfect_numbers(int start, int end) { int i, j, sum; for (i = start; i <= end; i++) { sum = 0; for (j = 1; j <= i / 2; j++) { if (i % j == 0) { sum += j; } } if (sum == i) { printf("%d is a perfect number.\n", i); } } } int main() { int lower, upper; printf("Enter range (lower and upper): "); scanf("%d%d", &lower, &upper); find_perfect_numbers(lower, upper); return 0; } ``` --- ### 打印特定图案 以下是一个打印三角形图案的示例： ```c #include <stdio.h> void print_pattern(int height) { for (int i = 1; i <= height; i++) { for (int j = 1; j <= i; j++) { printf("* "); } printf("\n"); } } int main() { int h; printf("Enter pattern height: "); scanf("%d", &h); print_pattern(h); return 0; } ``` 相关问题

### **题目1** - 判断三角形及计算面积 以下是Python实现： ```python import math def is_triangle(a, b, c): return a + b > c and a + c > b and b + c > a while True: try: a = float(input("请输入第一条边长：")) b = float(input("请输入第二条边长：")) c = float(input("请输入第三条边长：")) if not (isinstance(a, (int, float)) and isinstance(b, (int, float)) and isinstance(c, (int, float))): raise ValueError if is_triangle(a, b, c): p = (a + b + c) / 2 area = math.sqrt(p * (p - a) * (p - b) * (p - c)) print(f"可以构成三角形，面积为{area:.2f}") else: print("无法构成三角形") break except Exception as e: print("输入错误，请输入有效的数字！") ``` --- ### **题目2** - 回文数判断 以下是Python实现： ```python def is_palindrome(num): str_num = str(num) return str_num == str_num[::-1] num = int(input("请输入一个自然数：")) if is_palindrome(num): print(f"{num}是回文数") else: print(f"{num}不是回文数") ``` --- ### **题目3** - 数列之和 以下是Python实现： ```python def sum_of_sequence(a, n): total_sum = 0 current_term = "" for _ in range(n): current_term += str(a) total_sum += int(current_term) return total_sum try: a = int(input("请输入单个数字a（1≤a≤9）：")) n = int(input("请输入非负整数n（0≤n≤100000）：")) result = sum_of_sequence(a, n) print(f"结果为 {result}") except ValueError: print("输入有误，请重新运行程序并输入正确的值！") ``` --- ### **题目4** - 打印水仙花数 以下是Python实现： ```python # 输出所有三位水仙花数 for num in range(100, 1000): digits = [int(d) for d in str(num)] if num == sum([d ** 3 for d in digits]): print(num) print("\n四位水仙花数:") # 输出所有四位水仙花数 for num in range(1000, 10000): digits = [int(d) for d in str(num)] if num == sum([d ** 4 for d in digits]): print(num) ``` --- ### **题目5** - 判断闰年 以下是Python实现： ```python year = int(input("请输入一个年份：")) if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): print(f"{year} 年是闰年") else: print(f"{year} 年是平年") ``` --- ### **题目6** - 筛选素数 以下是Python实现： ```python def is_prime(num): if num < 2: return False for i in range(2, int(math.sqrt(num)) + 1): if num % i == 0: return False return True primes = [] count = 0 for i in range(2, 301): if is_prime(i): primes.append(i) count += 1 print("300以内的素数:", primes) print("素数总数:", count) ``` --- ###