实验三

实验一

#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：将分数转化为相应的等级。int，char。

问题二：缺少break，双引号改为单引号。

实验二

#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;
}

问题一：计算整数各位数字之和。

问题二：能。一个是迭代，一个是递归。

实验三

#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;
    }
}

问题一：进行幂运算。

问题二：是。

                      {1      n=0          

power（x,n)={x*power(x,n-1)     n为奇

             {【power（x,n/2)】²    n为偶

实验四

#include <stdio.h>

int classify_triangle(int a, int b, int c);

int main() {
    int a, b, c;
    while (scanf("%d %d %d", &a, &b, &c) != EOF) {
        int type = classify_triangle(a, b, c);
        switch (type) {
            case 0: printf("不能构成三角形\n"); break;
            case 1: printf("普通三角形\n"); break;
            case 2: printf("等边三角形\n"); break;
            case 3: printf("等腰三角形\n"); break;
            case 4: printf("直角三角形\n"); break;
        }
    }
    return 0;
}

int classify_triangle(int a, int b, int c) {
    if (a + b <= c || a + c <= b || b + c <= a) {
        return 0; 
    }
    if (a == b && b == c) {
        return 2;
    }
    if (a == b || a == c || b == c) {
        return 3;
    }
    if (a * a + b * b == c * c || a * a + c * c == b * b || b * b + c * c == a * a) {
        return 4;
    }
    else 
    return 1;
}

实验五

#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 x = 1;
    int y = 1;
    int i;
    
    for(i = 0; i < m; i++){
        x *=(n-i);
        y *=(m-i);
    }
    
    return x/y;
}

#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==0||m==n)
        return 1;
    if(m>n)
        return 0;
        
    return func(n-1,m)+func(n-1,m-1);
}

实验六

#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;
    int i;
    
    min = a;
    if (b < min) min = b;
    if (c < min) min = c;

    for (i = min; i >= 1; i--) {
        if (a % i == 0 && b % i == 0 && c % i == 0) {
            return i; 
        }
    }
}

实验七

#include <stdio.h>
#include <stdlib.h>
void print_charman(int n);
int main() {
    int n;
    printf("Enter n: ");
    while(scanf("%d", &n) != EOF){
        printf("input n: %d\n", n); 
        print_charman(n);           
        printf("\nEnter n: ");   
    }
    return 0;
}
void print_charman(int n) {   
    for (int i = 1; i <= n; i++) {      
        for (int j = 1; j < i; j++) {
            printf("\t"); 
        }
        for (int j = 1; j <= 2 * (n - i) + 1; j++) {
            printf("  O  \t"); 
        }
        printf("\n");              
        for (int j = 1; j < i; j++) {
            printf("\t");
        }
        for (int j = 1; j <= 2 * (n - i) + 1; j++) {
            printf(" <H> \t");
        }
        printf("\n");               
        for (int j = 1; j < i; j++) {
            printf("\t");
        }
        for (int j = 1; j <= 2 * (n - i) + 1; j++) {
            printf(" I I \t"); 
        }
        printf("\n");
    }
}