实验3
任务1
1 #include<stdio.h> 2 char score_to_grade(int score); 3 int main(){ 4 int score; 5 char grade; 6 while(scanf("%d",&score)!=EOF){ 7 grade = score_to_grade(score); 8 printf("分数：%d,等级：%c\n\n",score,grade); 9 10 } 11 return 0; 12 } 13 char score_to_grade(int score){ 14 char ans; 15 switch(score/10){ 16 case 10: 17 case 9: ans = 'A';break; 18 case 8: ans = 'B';break; 19 case 7: ans = 'C';break; 20 case 6: ans = 'D';break; 21 default: ans = 'E'; 22 } 23 return ans; 24 }
问题1： 函数score_to_grade功能：按照分数范围给分数打上等级
问题2：若无break,则只能在成绩为90-100分时输出ABCDE，其他不输出
任务2
1 #include<stdio.h> 2 int sum_digits(int n); 3 int main(){ 4 int n; 5 int ans; 6 while(printf("Enter n:"),scanf("%d",&n)!=EOF){ 7 ans = sum_digits(n); 8 printf("n=%d,ans=%d\n\n",n,ans); 9 } 10 return 0; 11 } 12 int sum_digits(int n){ 13 int ans = 0; 14 while(n!=0){ 15 ans+=n%10; 16 n/=10; 17 18 } 19 return ans; 20 }
问题1：求数字所有位上的数字之和
问题2：能实现同等效果，原代码为循环结构，改后为判断结构
任务3
1 #include <stdio.h> 2 int power(int x, int n); // 函数声明 3 int main() { 4 int x, n; 5 int ans; 6 while (printf("Enter x and n: "), scanf_s("%d%d", &x, &n) != EOF) { 7 ans = power(x, n); // 函数调用 8 printf("n = %d, ans = %d\n\n", n, ans); 9 } 10 return 0; 11 } 12 // 函数定义 13 int power(int x, int n) { 14 int t; 15 if (n == 0) 16 return 1; 17 else if (n % 2) 18 return x * power(x, n - 1); 19 else { 20 t = power(x, n / 2); 21 return t * t; 22 } 23 }
问题1：函数power的作用为计算x的n次幂
问题2：是递归函数 n=0时x^n=0,n>0时x^n=x*x^(n-1)
任务4
1 #include <stdio.h> 2 int is_prime(int n); 3 int main() { 4 int all = 0; 5 for (int num = 1; num < 100; num++) { 6 if (is_prime(num) && is_prime(num + 2)) { 7 printf("(%d, %d)\n", num, num + 2); 8 all++; 9 } 10 } 11 printf("100 以内的孪生素数总数为：%d\n", all); 12 return 0; 13 } 14 int is_prime(int n) 15 { 16 if (n <= 1) { 17 return 0; 18 } 19 if (n <= 3) { 20 return 1; 21 } 22 if (n % 2 == 0 || n % 3 == 0) { 23 return 0; 24 } 25 int i = 5; 26 while (i * i <= n) { 27 if (n % i == 0 || n % (i + 2) == 0) { 28 return 0; 29 } 30 i += 6; 31 } 32 return 1; 33 }
任务5
1 #include <stdio.h> 2 int move(int n, char from, char to); 3 int hanoi(int n, char A, char C, char B); 4 int num=0; 5 int main() { 6 int n; 7 8 while (scanf_s("%d", &n) != EOF) { 9 hanoi(n, 'A', 'C', 'B'); 10 printf("一共移动了%d次\n",num ); 11 } 12 return 0; 13 } 14 int move(int n, char from, char to) { 15 printf("%d: %c --> %c\n", n, from, to); num++; 16 } 17 int hanoi(int n, char A, char C, char B) { 18 if (n == 1) { 19 move(1, A, C); 20 21 return; 22 } 23 else { 24 hanoi(n - 1, A, B, C); 25 move(n, A, C); 26 hanoi(n - 1, B, C, A); 27 } 28 }
任务6
1 #include<stdio.h> 2 int func(int n, int m); 3 int main() 4 { 5 int n, m; 6 int ans; 7 while (scanf_s("%d%d", &n, &m) != EOF) 8 { 9 ans = func(n, m); 10 printf("n = %d,m = %d,ans = %d \n\n", n, m, ans); 11 12 } 13 14 return 0; 15 } 16 int func(int n, int m) 17 { 18 if (m == 0 || m == n) 19 return 1; 20 else if (n < m) 21 return 0; 22 else 23 return func(n - 1, m) + func(n - 1, m - 1); 24 }
任务7
#include <stdio.h> void print_charman(int n); int main() { int n; printf("Enter n: "); scanf_s("%d", &n); print_charman(n); return 0; } void print_charman(int n) { int i; int t = 0; for ( i = n; i >= 1; i--) { for (int j = 0; j < t; j++) printf("\t"); for (int j = 0; j < 2 * i - 1; j++) printf(" O\t"); printf("\n"); for (int j = 0; j < t; j++) printf("\t"); for (int j = 0; j < 2 * i - 1; j++) printf("<H>\t"); printf("\n"); for (int j = 0; j < t; j++) printf("\t"); for (int j = 0; j < 2 * i - 1; j++) printf("I I\t"); printf("\n"); t++; } }
