实验三
任务1
task1.1 源代码
#include<stdio.h> long long fac(int n); int main() { int i, n; printf("Enter n: "); scanf_s("%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; p = p * n; return p; }
修改过一行的代码
#include<stdio.h> long long fac(int n); int main() { int i, n; printf("Enter n: "); scanf_s("%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; }
运行效果
task 1.2
#include<stdio.h> int func(int, int); int main() { int k = 4, m = 1, p1, p2; p1 = func(k, m); // 函数调用 p2 = func(k, m);// 函数调用 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); }
static变量不会随着函数的结束而被释放
static变量不会被再次赋初值
任务2
迭代版
#include<stdio.h> void printSymbol(int n, char symbol) { for(int i=0;i<n;i++) { printf("%c", symbol); } } int main() { int n; char symbol; while (scanf_s("%d %c", &n, &symbol) != EOF) { printSymbol(n, symbol); // 函数调用 printf("\n"); } return 0; }
递归版
#include<stdio.h> void printSymbol(int n, char symbol) { if (n == 0)return; printSymbol(n - 1, symbol); printf("%c", symbol); } int main() { int n; char symbol; while (scanf_s("%d %c", &n, &symbol) != EOF) { printSymbol(n, symbol); // 函数调用 printf("\n"); } return 0; }
我认为此题迭代更好
因为迭代的思维方式在此题中更加直观易懂,且代码量也和递归方式相差不大
任务3
#include<stdio.h> long long fun(int n) { if (n == 0)return 0; else { return 2 * fun(n - 1)+1; } } int main() { int n; long long f; while (scanf_s("%d", &n) != EOF) { f = fun(n); // 函数调用 printf("n = %d, f = %lld\n", n, f); } return 0; }
任务4
#include<stdio.h> #include<math.h> int isPrime(int n) { for(int i=2;i<=sqrt(n);i++) { if (n % i == 0)return 0; } return 1; } int main() { int sum = 0; for(int i=101;i<=200;i++) { if(!isPrime(i)) { printf("%6d", i); sum++; if (sum % 10 == 0)printf("\n"); } } printf("\n一共有%d个非素数", sum); return 0; }
任务5
#include<stdio.h> #include<math.h> long fun(long s) { long a=0, b=1,num=0;//a用来存放每位的数,b用来记位数 while(s>0) { a = s % 10; if(a%2!=0) { num += a * b; b *= 10; } s /= 10; } return num; } int main() { long s, t; printf("Enter a number: "); while (scanf_s("%ld", &s) != EOF) { t = fun(s); // 函数调用 printf("new number is: %ld\n\n", t); printf("Enter a number: "); } return 0; }
任务6
#include<stdio.h> #include<math.h> double jie_cheng(int x) { if (x == 1)return 1; else { return x * jie_cheng(x - 1); } } double fun(int n) { double sum=0; for(int i=1;i<=n;i++) { if (i % 2 == 0)sum -= (1 / jie_cheng(i)); else sum += (1 / jie_cheng(i)); } return sum; } int main() { int n; double s; printf("Enter n(1~10): "); while (scanf_s("%d", &n) != EOF) { s = fun(n); // 函数调用 printf("n = %d, s= %f\n\n", n, s); printf("Enter n(1~10): "); } return 0; }