实验3

实验任务1

#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;
}

问题1：判断分数的等级；形参类型：int；返回值类型：char

问题2：没有break，最后结果都会变成'E'；"A"为字符串，应该用'A'

实验任务2

 

#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;
}

问题1：计算出输入数的各数位的数字之和

问题2：能

第一张为迭代，第二种为递归

实验任务3

 

#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;
    }
}

问题1：计算x的n次方

问题2：是

递归模式：power(x,0)=1;power(x,n(为奇数))=power(x,n-1);power(x,n(n为偶数))=power(x,n/2)

实验任务4

#include <stdio.h>
int is_prime(int n);
int main() {
    int count = 0;
    printf("100以内的学生素数：\n");
    for (int n = 2; n <= 98; n++) { // n+2 <= 100
        if (is_prime(n) && is_prime(n + 2)) {
            printf("%d %d\n", n, n + 2);
            count++;
        }
    }
    printf("\n100以内的学生素数共有%d个。\n", count);
    return 0;
}
int is_prime(int n) {
    if (n <= 1) {
        return 0;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}

实验任务5

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) {
    long long c=1;
    for(int i=1;i<=m;i++){
        c=c*(n-i+1)/i;
    }
    return (int)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(m==n||m==0)
        return 1;
    else
        return func(n-1,m)+func(n-1,m-1);
}

实验任务6

#include<stdio.h>
int gcd(int a,int b,int 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(int a, int b, int c) {
    int min=a;
    if(b<min)min=b;
    if(c<min)min=c;
    for(int i=min;i>=1;i--){
        if (a%i==0&&b%i==0&&c%i==0){
            return i;
        }
    }
    return 1;
}

实验任务7

 

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