实验3

task1

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#include <stdio.h>char score_to_grade(int score);  // 函数声明int main() {int score;char grade;while(scanf("%d", &score) != EOF) {grade = score_to_grade(score);  // 函数调用printf("分数: %d, 等级: %c\n\n", score, grade);}return 0;
}// 函数定义
char score_to_grade(int score) {char ans;switch(score/10) {case 10:case 9:   ans = 'A'; break;case 8:   ans = 'B'; break;case 7:   ans = 'C'; break;case 6:   ans = 'D'; break;default:  ans = 'E';}return ans;
}


将分数转化为等级
型参是整形，返回字符

没加break

task2

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#include <stdio.h>int sum_digits(int n);  // 函数声明int main() {int n;int ans;while(printf("Enter n: "), scanf("%d", &n) != EOF) {ans = sum_digits(n);    // 函数调用printf("n = %d, ans = %d\n\n", n, ans);}return 0;
}// 函数定义
int sum_digits(int n) {int ans = 0;while(n != 0) {ans += n % 10;n /= 10;}return ans;
}


求各位数的和
能，是递归的思想

task 3

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#include <stdio.h>int power(int x, int n);    // 函数声明int main() {int x, n;int ans;while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {ans = power(x, n);  // 函数调用printf("n = %d, ans = %d\n\n", n, ans);}return 0;
}// 函数定义
int power(int x, int n) {int t;if(n == 0)return 1;else if(n % 2)return x * power(x, n-1);else {t = power(x, n/2);return t*t;}
}

求x的n次幂

准备用笔写，如果没有删除这行记得踢我一下

============================================================================
task 4

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#include<stdio.h>bool is_prime(int);int main()
{for(int i = 1 ; i <= 100 ; i++){if(is_prime(i) && is_prime(i+2))printf("%d %d\n",i,i+2);}return 0;
}bool is_prime(int x)
{if(x==1)return 0;for(int i=2;i<=x/2;i++){if(x % i == 0)return 0;}return 1;
}

 =============================================================================================== task 5

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#include<stdio.h>void Hanoi(int n , char from , char to ,char temp);int cnt;int main()
{int n;scanf("%d",&n);Hanoi(n , 'A' , 'C' , 'B');printf("%d",cnt);return 0;
}void Hanoi(int n , char from , char to ,char temp)
{if (n==1){printf("%d : %c --> %c\n",n , from , to);cnt++;}else{cnt++;Hanoi(n-1,from,temp,to);printf("%d : %c --> %c\n",n , from , to);Hanoi(n-1,temp,to,from);}
}

 ============================================================================================== task 6.1

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#include <stdio.h>
int func(int n, int m);   // 函数声明int main() {int n, m;int ans;while(scanf("%d%d", &n, &m) != EOF) {ans = func(n, m);   // 函数调用printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);}return 0;
}// 函数定义
// 待补足。。。（分别用迭代和递归实现）
int func(int n, int m)
{int mu=1,zi=1;int i;for( i=1; i<= m ;i++){zi*=n-i+1;mu*=i;}return zi/mu;
}


task 6.2

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#include <stdio.h>
int func(int n, int m);   // 函数声明int main() {int n, m;int ans;while(scanf("%d%d", &n, &m) != EOF) {ans = func(n, m);   // 函数调用printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);}return 0;
}// 函数定义
// 待补足。。。（分别用迭代和递归实现）
int func(int n, int m)
{if(m>n)return 0;if(m==0)return 1;if(m>0){return func(n-1,m)+func(n-1,m-1);}}

task 7

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#include <stdio.h>
#include <stdlib.h>void print_charman(int n);int main() 
{int n;printf("Enter n: ");scanf("%d", &n);print_charman(n); // 函数调用return 0;
}
void print_charman(int n)
{int i,j;for( i = 0 ; i<n ; i++){for( j = 0 ; j<i ; j++){printf("\t");}for( j = 0 ; j<n-i ; j++){printf(" O \t");}for( j = 0 ; j<n-i-1 ; j++){printf(" O \t");}printf("\n");for( j = 0 ; j<i ; j++){printf("\t");}for( j = 0 ; j<n-i ; j++){printf("<H>\t");}for( j = 0 ; j<n-i-1 ; j++){printf("<H>\t");}printf("\n");for( j = 0 ; j<i ; j++){printf("\t");}for( j = 0 ; j<n-i ; j++){printf("I I\t");}for( j = 0 ; j<n-i-1 ; j++){printf("I I\t");}printf("\n\n");}}