实验三

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;
}

 问题一：把分数转换为等级；整型和字符型；字符型

问题二：程序无法终止，一直输出该分数对应的等级和以下的等级（如case 8输出BCDE）

 

 

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

 问题一：输入的数每一位的和

问题二：能    一个是递归，一个是迭代

 

 

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;
    }
}

 

 问题一：算x的n次方

问题二：是

 

 

 

 

task4

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

 

 

task5

1.迭代法

#include<stdio.h>
int func(int n, int m) {
    int ans,i,j,a=1,b=1;
    if (n < m)
        ans = 0;
    else {
        for (i = 1; i <= m; i++)
            a *= i;
        for (j = n; j > n - m; j--)
            b *= j;
        ans = b / a;
    }
    return ans;
}
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;
}

 

 2.递归法

#include<stdio.h>
int func(int n, int m) {
     if (n < m)
       return 0;
     if (m == 0||m==n)
       return 1;
return func(n-1, m)+func(n-1,m-1);
}
int main() {
   int n, m;
   int ans;
   while (scanf_s("%d%d", &n, &m) != EOF) {
   ans = func(n, m);
   printf("n=%d,m=%d,ans=%d\n\n", n, m, ans);
}
return 0;
}

 

 

 

task6

#include<stdio.h>
int gcd(int a, int b, int c);
int min(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 m, i, a1, b1, c1, x, y;
    x = min(a, b, c);
    m = min(a, b, c);
    for (i = m; i > 0; i--) {
        a1 = a % i;
        b1 = b % i;
        c1 = c % i;
        if (a1 + b1 + c1 == 0)
            return i;
    }
}
int min(int a, int b, int c) {
    if (a <= b && a <= c)
        return a;
    else if (b <= a && b <= c)
        return b;
    else
        return c;
}

 

 task7

 #define _CRT_SECURE_NO_WARNINGS
 #include <stdio.h>
 #include <stdlib.h>
 void print_charman(int n);
 void ans1(int m, int indent);
 int main() {
         int n;
         printf("Enter n: ");
         scanf("%d", &n);
         print_charman(n); // 函数调用
             return 0;
}

 void print_charman(int n) {
            for (int i = 0; i < n; i++) {
                 int m = 2 * (n - i) - 1;
             int indent = i;
                 ans1(m, indent);
                 printf("\n");
    }
}
 void ans1(int m, int indent) {
     int i, k;
     for (k = 0; k < indent; k++) printf("\t");
     for (i = 0; i < m; i++) {
         printf(" O \t");
     }
     printf("\n");
     for (k = 0; k < indent; k++) printf("\t");
     for (i = 0; i < m; i++) {
         printf("<H>\t");
     }
     printf("\n");
     for (k = 0; k < indent; k++) printf("\t");
     for (i = 0; i < m; i++) {
         printf("I I\t");
     }
     printf("\n");
 }