实验3

#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：实现分数对应等级，形参是整形，返回字符型

问题2：没有break，选中后会从此条往下全部运行一遍；单个字符应用‘’

 

#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;
}

问题1：求n各位数字的总和

问题2：能，一种用循环解决，另一种用递归解决，依次识别最后一位

#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;
}
}

问题1：power为次方计算

问题2：是递归模型f(n-1)=f(n)x,n是单数, f(n/2)=f(n)*f(n),n是双数

 

#include<stdio.h>
int is_prime(int x);
int main() {
    int s=0,a=0,b,n=0;
    for (int i = 3; i < 100; i = i + 2) {
        b = a;
        if (is_prime(i) == 1) {
            a = i;
            s++;
        }
        else {
        s = 0;
    }
        if (s >= 2) {
            printf("%d  %d\n", b, a);
            n++;
        }
    }
    printf("共%d对素数",n);
    return 0;
}
int is_prime(int x) {
    for (int i = 2; i < x; i++) {
        if (x % i == 0)
            return 0;
    }
    return 1;
}

#include <stdio.h>
int func(int n, int m); // 函数声明
int main() {
    int n, m;
    int ans;
    while (scanf_s("%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 a=1, b=1, c=1;
    for (int i = n; i > 0; i--) {
        a *= i;
    }
    for (int i = m; i > 0; i--) {
        b *= i;
    }
    for (int i =n-m; i > 0; i--) {
        c *= i;
    }
    return a / (b * c);
}

#include <stdio.h>
int func(int n, int m); // 函数声明
int main() {
    int n, m;
    int ans;
    while (scanf_s("%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 (m > n)
        return 0;
    else if (n == m||n==0||m==0)
        return 1;
    else
        return func(n - 1, m) + func(n - 1, m - 1);
}

#include <stdio.h>
int gcd(int a,int b,int c);
int main() {
    int a, b, c;
    int ans;
    while (scanf_s("%d%d%d", &a, &b, &c) != EOF) {
        ans = gcd(a, b, c); // 函数调用
        printf("最大公约数: %d\n\n", ans);
    }
    return 0;
}
int gcd(int a, int b, int c) {
    int i;
    for (i = a; i > b && i > c; i--);
    for (i; i > 0; i--) {
        if (a % i == 0 && b % i == 0 && c % i == 0)
            return i;
    }
}

#include <stdio.h>
#include <stdlib.h>
int print_charman(int n);
int main() {
    int n;
    printf("Enter n: ");
    scanf_s("%d", &n);
    print_charman(n); // 函数调用
    return 0;
}
int print_charman(int n){
    int t = 2*n-1;
    for (t; t > 0;t-=2) {
        for (int i=(2*n-1-t)/2;i>0;i--) {
            printf("\t");
        }
        for (int i=t; i > 0;i--) {
            printf(" O \t");
        }
        printf("\n");
        for (int i = (2 * n - 1 - t) / 2; i > 0; i--) {
            printf("\t");
        }
        for (int i = t; i > 0; i--) {
            printf("<H>\t");
        }
        printf("\n");
        for (int i = (2 * n - 1 - t) / 2; i > 0; i--) {
            printf("\t");
        }
        for (int i = t; i > 0; i--) {
            printf("I I\t");
        }
        printf("\n");

    }
}