实验3
task1源代码：
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("%d", &score) != EOF) { 10 grade = score_to_grade(score); 11 printf("分数: %d, 等级: %c\n\n", score, grade); 12 } 13 14 return 0; 15 } 16 char score_to_grade(int score) { 17 char ans; 18 19 switch(score/10) { 20 case 10: 21 case 9: ans = 'A'; break; 22 case 8: ans = 'B'; break; 23 case 7: ans = 'C'; break; 24 case 6: ans = 'D'; break; 25 default: ans = 'E'; 26 }
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问题１：功能是将分数转化为评级，形参为整型，返回值为字符型
问题２：有，在处理完数据后不会跳出switch语句，继续判断下一个case
task2源代码：
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("%d", &n) != EOF) { 10 ans = sum_digits(n); 11 printf("n = %d, ans = %d\n\n", n, ans); 12 } 13 14 return 0; 15 } 16 int sum_digits(int n) { 17 int ans = 0; 18 19 while(n != 0) { 20 ans += n % 10; 21 n /= 10; 22 } 23 24 return ans; 25 }
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问题1：功能是实现每位上的数字相加
问题2：可以，法2使用了递归的思想
task3源代码:
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("%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 int power(int x, int n) { 17 int t; 18 19 if(n == 0) 20 return 1; 21 else if(n % 2) 22 return x * power(x, n-1); 23 else { 24 t = power(x, n/2); 25 return t*t; 26 } 27 }
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问题1：实现x的n次方的计算
问题2：是递归函数， 指数如果是奇数则为x**n*(x**n-1),如果是偶数则为x**(n/2)*x**(n/2)
task4源代码：
1 #include <stdio.h> 2 #include <math.h> 3 int main() 4 { 5 int is_prime(int); 6 printf("100以内的孪生素数：\n"); 7 int i,s=0; 8 for (i = 2;i<=100;i++){ 9 if (is_prime(i)&&is_prime(i+2)){ 10 s=s+1; 11 printf("%d %d\n",i,i+2); 12 } 13 } 14 printf("100以内的孪生素数共有%d个\n",s); 15 return 0; 16 } 17 int is_prime(int n){ 18 int i,m; 19 m = sqrt(1.0*n); 20 for (i = 2;i <= m;i++){ 21 if(n % i == 0)break; 22 } 23 if (i>m && n>=2) 24 return 1; 25 else 26 return 0; 27 }
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实验5源代码：
1 #include<stdio.h> 2 #include<stdlib.h> 3 void hanoi(unsigned int n,char from,char temp,char to); 4 void moveplate(unsigned int n,char from,char to); 5 int s = 0; 6 int main() 7 { 8 unsigned int n; 9 while(scanf("%u",&n) != EOF){ 10 s=0; 11 hanoi(n,'A','B','C'); 12 printf("一共移动了%d次\n",s); 13 } 14 system("pause"); 15 return 0; 16 } 17 void hanoi(unsigned int n,char from,char temp,char to){ 18 if(n == 1) 19 moveplate(n,from,to); 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(unsigned int n,char from,char to){ 27 printf("%u:%c-->%c\n",n,from,to); 28 s = s + 1; 29 }
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task6源代码1:
1 #include<stdio.h> 2 int main() 3 { 4 int i,h,n,m,up=1,down=1,ans; 5 while(scanf("%d%d",&n,&m) != EOF){ 6 if(m == 0) 7 ans = 1; 8 else{ 9 for(i=n;i >= n-m+1;i--){ 10 up = up * i; 11 } 12 for(h=m;h >= 1;h--){ 13 down = down * h; 14 } 15 ans = up/down; 16 } 17 printf("n = %d,m = %d,ans = %d\n",n,m,ans); 18 } 19 return 0; 20 }
源代码2:
1 #include <stdio.h> 2 int func(int n,int m); 3 int power(int n); 4 int main(){ 5 int n,m,ans; 6 while(scanf("%d%d",&n,&m)!=EOF){ 7 ans=func(n,m); 8 printf("n=%d,m=%d,ans=%d\n\n",n,m,ans); 9 } 10 return 0; 11 } 12 int func(int n,int m){ 13 if(m>n) 14 return 0; 15 else if(n==m||m==0) 16 return 1; 17 else 18 return func(n-1,m)+func(n-1,m-1); 19 }
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task7源代码：
1 #include <stdio.h> 2 #include <stdlib.h> 3 4 void print_charman(int n); 5 int main() { 6 int n; 7 8 printf("Enter n: "); 9 scanf("%d", &n); 10 print_charman(n); 11 12 return 0; 13 } 14 void print_charman(int n){ 15 int m = ((n*2)-1) ,count=0,i,x,b,a; 16 while(m!=-1){ 17 for(x=0;x<count;x++){ 18 printf("\t"); 19 } 20 for(i=0; i<m ;i++){ 21 printf(" O\t"); 22 } 23 printf("\n"); 24 for(x=0;x<count;x++){ 25 printf("\t"); 26 } 27 for(a=0; a<m ;a++){ 28 printf("<H>\t"); 29 } 30 printf("\n"); 31 for(x=0;x<count;x++){ 32 printf("\t"); 33 } 34 for(b=0; b<m ;b++){ 35 printf("I I\t"); 36 } 37 printf("\n"); 38 count++; 39 m=m-2; 40 } 41 }
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