实验3

task1

#include <stdio.h> 
char score_to_grade(int score);
int main(){
    int score;
    char grade;
    while(scanf("%d",&score)!=EOF){
        grade=score_to_grade(score);
        printf("分数:%d,等级:%c\n\n",score,grade);
        }
        return 0;
}
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;
}

A1：将学生分数转化成对应等级；int；char

A2：没有break 导致逻辑错误

task2

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

A1：计算输入数字各个位数字之和

A2：能，第一种是迭代法 ，第二种是递归法

task3

#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;
    }
}

 A1：计算x的n次方

A2：

task4

#include <stdio.h>
int is_prime(int n) {
    if (n < 2) {
        return 0;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}

int main() {
    int count = 0;
    printf("100以内的孪生素数：\n");
    for (int n = 2; n + 2 <= 100; n++) {
        if (is_prime(n) && is_prime(n + 2)) {
            printf("%d %d\n", n, n + 2);
            count++;
        }
    }
    printf("100以内的孪生素数共有%d个。\n", count);
    return 0;
}

 

task5

 迭代

#include <stdio.h>
int func(int n, int m);  

int main() {
    int n, m;
    int ans;

    while(scanf("%d%d", &n, &m) != EOF) {
        ans = func(n, m);   
        printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
    }
        
    return 0;
}
int func(int n, int m){
    int a=1,b=1,c,i,j;
    for(i=n-m+1;i<=n;i++){
        a*=i;
    }
    for(j=1;j<=m;j++){
        b*=j;
    }
    c=a/b;
    return c;
    
}

递归

#include <stdio.h>
int func(int n, int m);  

int main() {
    int n, m;
    int ans;

    while(scanf("%d%d", &n, &m) != EOF) {
        ans = func(n, m); 
        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{
        return func(n-1,m)*func(n-1,m-1);
    }
}

task6

#include <stdio.h>

int gcd(a,b,c); 

int main() {
    int a, b, c;
    int ans;

    while(scanf("%d%d%d", &a, &b, &c) != EOF) {
        ans = gcd(a, b, c);  
        printf("最大公约数: %d\n\n", ans);
    }

    return 0;
}
int gcd(a,b,c){
    int m;
    m=(a<b?a:b)<c?(a<b?a:b):c;
    if(a%m==0&&b%m==0&&c%m==0){
        return m; 
    }
    else{
    int i;
        for(i=m;i>0;i--){
            if(a%i==0&&b%i==0&&c%i==0){
                return i;
            }
        }
    }
        
    
}

 

task7

#include <stdio.h>
#include <stdlib.h>

void print_charman(int n); 
int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); 
       
    return 0;
}
void print_charman(int n){
int i,m,row,j;
for(i=0;i<n;i++){

    for(row=0;row<3;row++){
            for(m=0;m<i;m++){
        printf("    \t");
    }
        for(j=0;j<2*n-1-i*2;j++){
            
            switch(row){
                case 0:printf(" o\t");continue;
                case 1:printf("<H>\t");continue;
                case 2:printf("I I\t");continue;
            }
            printf("\n");
        }
            printf("\n");
    }
} 
}