实验三

任务一

#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;
}

1.对输入的分数进行评级；整型；字符型

2.无break，switch语句将不会终止

任务二

#include <stdio.h>
int sum_digits(int n); 
int main() {
    int n;
    int ans;
    while(printf("Enter n: "), scanf("%d", &n) != EOF) {
        ans = sum_digits(n);    
        printf("n = %d, ans = %d\n\n", n, ans);
    }

    return 0;
}
int sum_digits(int n) {
    int ans = 0;
    while(n != 0) {
        ans += n % 10;
        n /= 10;
    }
    return ans;
}

1.求各位上数的和

2.能，前者迭代，后者递归

任务三

#include <stdio.h>
int power(int x, int n);    
int main() {
    int x, n;
    int ans;
    while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {
        ans = power(x, n);  
        printf("n = %d, ans = %d\n\n", n, ans);
    }
    return 0;
}
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;
    }
}

1.计算x的n次方

2.能

任务四

#include <stdio.h>
#include <math.h> 
int is_prime(int n); 
int main() {
   int n;
   int cnt = 0;
   printf("100以内的孪生素数:\n");
   for (n = 1; n <= 98; n++) {
        if (is_prime(n) && is_prime(n + 2))  {
            printf("%d %d\n", n, n + 2);
            cnt++;}
            }
    printf("孪生素数共有%d个\n", cnt);
    return 0;
 }
 int is_prime(int n)
 {
    int i;
    if (n == 1)
        return 0;
    if (n == 2) {
        return 1;   }
    if (n % 2 == 0)
        return 0;
    for (i = 3; i <= sqrt(n); i += 2){
        if (n % i == 0) {
             return 0;
             }
    }
    return 1;
 }

 任务五

#include<stdio.h>
#include<stdlib.h>
void hanoi(unsigned int n,char from,char temp,char to);
void moveplate(unsigned int n,char from,char to);
int sum=0;
int main()
{unsigned int n;
extern int sum;
while(scanf("%u",&n)!=EOF){
hanoi(n,'A','B','C');
printf("共移动%d次\n",sum);
}
system("pause");
return 0;
}
void hanoi(unsigned int n,char from,char temp,char to)
{;
if(n==1){
moveplate(n,from,to);}
else
{hanoi(n-1,from,to,temp);
moveplate(n,from,to);
hanoi(n-1,temp,from,to);
}
}
void moveplate(unsigned int n,char from,char to)
{extern int sum;
printf("%u:%c-->%c\n",n,from,to);
sum=sum+1;
}

任务六

#include <stdio.h>
int func(int n, int m);   // 函数声明
int fac(int n); 
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;
}
int func(int n,int m){
    int s;
    s=fac(n)/fac(m)/fac(n-m);
    if (n<m){
        return 0;
    }
    else{
        return s;
    }
}
int fac(int n){
    int i,s=1;
    if(n==0){
        return 1;
    }
    else{
        for(i=1;i<=n;i++){
            s=s*i;
        }
        return s;
    }
}

#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;
}
int func(int n,int m){
    if(n==m||m==0){    
    return 1;}
    else if(n<m)
    return 0;
    else{
        return func(n-1,m)+func(n-1,m-1);
    }
}

任务七

#include <stdio.h>
#include <stdlib.h>
int m;
void print_charman(int n);
int main() {
int n;
int m;
m=n;
printf("Enter n: ");
scanf("%d", &n);
print_charman(n);   
    return 0;
}
void print_charman(int n){
    int i;
    int m;
    if(n>0){
    for(i=0;i<(m-n)*5;i++){
        printf(" ");
    }
    for(i=0;i<n*2-1;i++){
        printf(" O   ");}
    printf("\n");
    for(i=0;i<(m-n)*5;i++){
        printf(" ");}
    for(i=0;i<n*2-1;i++){
        printf("<H>  ");}
    printf("\n");
    for(i=0;i<(m-n)*5;i++){
        printf(" ");}
    for(i=0;i<n*2-1;i++){
        printf("I I  ");}
    printf("\n");
    print_charman(n-1);}
    else
    {printf("\n");
    }
}