实验三
实验三
实验任务一
源代码
#include <stdio.h> #include <stdlib.h> #include <time.h> #include <windows.h> #define N 80 void print_text(int line, int col, char text[]); void print_spaces(int n); void print_blank_lines(int n); int main() { int line, col, i; char text[N] = "hi, April~"; srand(time(0)); for(i = 1; i <= 10; ++i) { line = rand() % 25; col = rand() % 80; print_text(line, col, text); Sleep(1000); } return 0; } void print_spaces(int n) { int i; for(i = 1; i <= n; ++i) printf(" "); } void print_blank_lines(int n) { int i; for(i = 1; i <= n; ++i) printf("\n"); } void print_text(int line, int col, char text[]) { print_blank_lines(line-1); print_spaces(col-1); printf("%s", text); }
实验结论
在0~24行,0~79个空格组成的空间中随即在一处地方打印”hi,April~“,并且重复打印10次。
实验任务二
task2-1
源代码
#include <stdio.h> long long fac(int n); int main() { int i, n; printf("Enter n: "); scanf("%d", &n); for (i = 1; i <= n; ++i) printf("%d! = %lld\n", i, fac(i)); return 0; } long long fac(int n) { static long long p = 1; printf("p = %lld\n", p); p = p * n; return p; }
实验结论
task2-2
源代码
#include <stdio.h> int func(int, int); int main() { int k = 4, m = 1, p1, p2; p1 = func(k, m); p2 = func(k, m); printf("%d, %d\n", p1, p2); return 0; } int func(int a, int b) { static int m = 0, i = 2; i += m + 1; m = i + a + b; return m; }
实验结论
实验思考
局部static变量的特性:在自定义函数中,static变量不会随着函数的调用而被重置。相反,系统分配给static变量的空间是固定的,不会收走。
实验任务三
源代码
#include <stdio.h> long long func(int n); int main() { int n; long long f; while (scanf_s("%d", &n) != EOF) { f = func(n); printf("n = %d, f = %lld\n", n, f); } return 0; }
long long func(int n) {
if (n == 0)
return 0;
else
return func(n - 1) * 2 + 1;
}
实验结论
实验任务四
源代码
//task4 #include <stdio.h> int func(int n, int m); int main() { int n, m; while (scanf_s("%d%d", &n, &m) != EOF) printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m)); return 0; } // 函数定义 int func(int n, int m) { if (m == n || m == 0) return 1; else if (m > n) return 0; else if (m == 1) return n; else return func(n - 1, m) + func(n - 1, m - 1); }
实验结论
实验任务五
task5-1
源代码
//task5_1 #include <stdio.h> double mypow(int x, int y); // 函数声明 int main() { int x, y; double ans; while (scanf_s("%d%d", &x, &y) != EOF) { ans = mypow(x, y); // 函数调用 printf("%d的%d次方: %g\n\n", x, y, ans); } return 0; } // 函数定义 迭代算法 double mypow(int x, int y) { int i; double sum = 1; if (y > 0) for (i = 1; i <= y; i++) sum = sum * x; else if (y < 0) for (i = 1; i <= -y; i++) sum = sum * 1 / (double)x; else if (y = 0) sum = x; }
实验结论
task 5-2
源代码
//task5_2 #include <stdio.h> double mypow(int x, int y); // 函数声明 int main() { int x, y; double ans; while (scanf_s("%d%d", &x, &y) != EOF) { ans = mypow(x, y); // 函数调用 printf("%d的%d次方: %g\n\n", x, y, ans); } return 0; } // 函数定义 递归函数 double mypow(int x, int y) { if (y == 0) return 1; else if (y > 0) return mypow(x, y-1) * x; else return mypow(x, y+1) / x; }
实验结论
实验任务六
源代码
//task6 #include<stdio.h> #include<stdlib.h> void hanoi(unsigned int n, char from, char temp, char to); void moveplate(unsigned int n, char from, char to); int i = 0; int main() { int n; while (scanf_s("%u", &n) != EOF) { i = 0; hanoi(n, 'A', 'B', 'C'); printf("一共移动了%u次\n", i); system("pause"); } return 0; } void hanoi(unsigned int n, char from, char temp, char to) { if (n == 1) moveplate(n, from, to); else { hanoi(n - 1, from, to, temp); moveplate(n, from, to); hanoi(n - 1, temp, from, to); } } void moveplate(unsigned int n, char from, char to) { printf("%u: %c-->%c\n", n, from, to); i++; }
实验结论
实验任务七
源代码
#include<stdio.h> #include<stdlib.h> int is_prime(int n); int main() { int i, j; for (i = 4; i <= 20; i += 2) for (j = 2; j <= i / 2; j++) { if (is_prime(j) && is_prime(i - j)) { printf("%d=%d+%d\n", i, j, i - j); break; } } return 0; } int is_prime(int n) { int i; for (i = 2; i < n; i++) { if (n % i == 0) return 0; } return 1; }
实验结论
实验任务八
源代码
#include <stdio.h> #include <math.h> long func(long s); // 函数声明 int main() { long s, t; printf("Enter a number: "); while (scanf_s("%ld", &s) != EOF) { t = func(s); // 函数调用 printf("new number is: %ld\n\n", t); printf("Enter a number: "); } return 0; } // 函数定义 // 待补足。。。 long func(long s) { int i,k=0,n=0; for (; s / 10 != 0; s = s / 10) { i = s % 10; if (i % 2 != 0) k = k * 10 + i; } if (s % 2 != 0) k = k * 10 + s; for (; k % 10 != 0; k=k / 10) n = n * 10 + k % 10; return n; }