实验3 c语言函数应用编程
task1:
1 #include<stdio.h> 2 3 char score_to_grade(int score); 4 5 int main() { 6 int score; 7 char grade; 8 9 while (scanf_s("%d",&score) != EOF) { 10 grade = score_to_grade(score); 11 printf("分数:%d,等级:%c\n\n", score, grade); 12 13 } 14 15 return 0; 16 17 } 18 19 char score_to_grade(int score) { 20 char ans; 21 22 switch (score / 10) { 23 case 10: 24 case 9: ans = 'A';break; 25 case 8: ans = 'B';break; 26 case 7: ans = 'C';break; 27 case 6: ans = 'D';break; 28 default: ans = 'E'; 29 30 } 31 return ans; 32 33 }
问题1：功能是将键盘输入的分数转化成对应的等级 形参类型是整型 返回值类型是字符型
问题2：1 改成双引号后无法输出对应的等级 2 语句后不加break,程序执行到对应语句后无法终止，还会输出后面语句的结果
task2:
1 #include<stdio.h> 2 3 int sum_digits(int n); 4 5 int main() { 6 int n; 7 int ans; 8 9 while (printf("Enter n:"), scanf_s("%d", &n) != EOF) { 10 ans = sum_digits(n); 11 printf("n=%d,ans%d\n\n", n, ans); 12 13 } 14 15 return 0; 16 } 17 18 19 int sum_digits(int n) { 20 int ans = 0; 21 while (n != 0) { 22 ans += n % 10; 23 n /= 10; 24 25 } 26 return ans; 27 }
问题1：功能是得出输入数各位的加和
问题2：可以实现同等效果 原算法将数字的各位取出依次加和，改后通过递归实现
task3:
1 #include<stdio.h> 2 3 int power(int x, int n); 4 5 int main() { 6 int x, n; 7 int ans; 8 9 while (printf("Enter x and n :"), scanf_s("%d%d", &x, &n) != EOF) { 10 ans = power(x, n); 11 printf("n=%d,ans=%d\n\n", n, ans); 12 13 } 14 return 0; 15 } 16 17 int power(int x, int n){ 18 int t; 19 20 if (n == 0) 21 return 1; 22 else if (n % 2) 23 return x * power(x, n - 1); 24 else { 25 t = power(x, n / 2); 26 return t * t; 27 28 } 29 30 }
问题1：功能是计算x的n次方
问题2：是递归函数
task4:
1 #include<stdio.h> 2 #include<math.h> 3 int is_prime(int n); 4 5 6 int main() { 7 int n,sum; 8 n = 1; 9 sum = 0; 10 printf("100以内的孪生素数：\n"); 11 for (n = 1;n <= 98;++n) { 12 if (is_prime(n)&& is_prime(n + 2)) { 13 printf("%d %d\n", n, n + 2); 14 sum = sum + 1; 15 } 16 } 17 printf("100以内的孪生素数共有%d个", sum); 18 19 return 0; 20 21 22 } 23 24 25 26 27 28 29 30 31 32 int is_prime(int n) { 33 int i, m; 34 m = sqrt(n); 35 36 for (i = 2;i<=m;++i) 37 if (n % i == 0) 38 break; 39 if (i > m&&n>1) 40 return 1; 41 else 42 return 0; 43 } 44
task5:
1 #include<stdio.h> 2 void hanoi(int n, char from, char temp, char to); 3 void moveplate(int n, char from, char to); 4 auto sum = 0; 5 int main() { 6 int n; 7 while (scanf_s("%d", &n)!=EOF) { 8 hanoi(n, 'A', 'B', 'C'); 9 printf("一共移动了%d次\n\n", sum); 10 sum = 0; 11 } 12 return 0; 13 } 14 15 void hanoi(int n, char from, char temp, char to) { 16 if (n == 1) { 17 moveplate(n, from, to); 18 } 19 20 else { 21 hanoi(n - 1, from, to,temp); 22 moveplate(n, from, to); 23 hanoi(n - 1, temp, from, to); 24 } 25 } 26 void moveplate(int n, char from, char to){ 27 printf("%d:%c-->%c\n", n, from, to); 28 sum += 1; 29 }
task6:
1 #include<stdio.h> 2 int func(int n, int m); 3 4 int main(){ 5 int n, m; 6 int ans; 7 8 while (scanf_s("%d%d", &n, &m) != EOF) { 9 ans = func(n, m); 10 printf("n=%d,m=%d,ans=%d\n\n", n, m, ans); 11 12 } 13 return 0; 14 } 15 16 int func(int n, int m) { 17 int x=1, y=1; 18 int i, j; 19 for (i = n;i >= (n - m + 1);--i) 20 x = x * i; 21 for (j = 1;j <= m;++j) 22 y = y * j; 23 return x / y; 24 25 26 27 28 29 }
1 #include<stdio.h> 2 int func(int n, int m); 3 4 int main(){ 5 int n, m; 6 int ans; 7 8 while (scanf_s("%d%d", &n, &m) != EOF) { 9 ans = func(n, m); 10 printf("n=%d,m=%d,ans=%d\n\n", n, m, ans); 11 12 } 13 return 0; 14 } 15 16 int func(int n, int m) { 17 if (m == 0 || n == m ) 18 return 1; 19 else if(m>n) 20 return 0; 21 else { 22 return func(n - 1, m) + func(n - 1, m - 1); 23 } 24
task7:
1 #include<stdio.h> 2 3 int main() { 4 int n; 5 printf("Enter n:"); 6 scanf_s("%d", &n); 7 print_charman(n); 8 9 return 0; 10 } 11 12 13 int print_charman(int n) { 14 int i,j; 15 for (i = n;i > 0;--i) { 16 for (int m = 0;m < n - i;m++) 17 printf("\t"); 18 for (j = 1;j <=(2*i-1);j++) { 19 printf(" o \t"); 20 } 21 printf("\n"); 22 23 for (int m = 0;m < n - i;m++) 24 printf("\t"); 25 for (j = 1;j <= (2 * i - 1);j++) { 26 printf("<H>\t"); 27 } 28 printf("\n"); 29 30 for (int m = 0;m < n - i;m++) 31 printf("\t"); 32 for (j = 1;j <= (2 * i - 1);j++) { 33 printf("I I\t"); 34 } 35 printf("\n"); 36 37 38 39 40 41 42 }
