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);
} r
eturn 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;
}

 

 

功能:根据分数评判等级，形参是整形，返回值是字符型

修改后的代码会将等级从A一直输出至最后，没有跳出语句

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;
}

 

函数的作用是求一个数各个位数上的数之和

改后的函数能实现相同的作用，这是一种递归效果，且有出口。原函数是迭代

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;
}
}

 

power的功能是实现求x的n次方

Task4

#include <stdio.h>
#include <math.h>
int is_prime(int n);
int main(){
    int count = 0;
    printf("100以内的孪生素数：\n");
    for(int n=3;n<99;n+=2){
        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",count);
    
    return 0;
}
int is_prime(int n){
    if(n<2){
        return 0;
    }
    for(int i=2;i<=sqrt(n);i++){
        if(n%i==0){
            return 0； 
        }
    } 
    return 1;
}

 

#include <stdio.h>
void hanoi(int n, char source, char target, char auxiliary, int* move_count);

int main() {
    int n;
    while (1) {
        printf("请输入盘子数量n：");
        if (scanf("%d", &n) != 1) {
            break;
        }
        int move_count = 0;
        printf("\n", n);
        hanoi(n, 'A', 'C', 'B', &move_count);
        printf("一共移动了 %d 次。\n", move_count);
    }
    return 0;
}

void hanoi(int n, char source, char target, char auxiliary, int* move_count) {
    if (n > 0) {
        hanoi(n - 1, source, auxiliary, target, move_count);
        printf("%d: %c --> %c\n", n, source, target);
        (*move_count)++;
        hanoi(n - 1, auxiliary, target, source, move_count);
    }
}

#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, m, ans);
    }
    return 0;
}

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

#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, m, ans);
    }
    return 0;
}
int func(int n, int m) {
    if (m > n) {
        return 0;
    }
    if (m == 0 || m == n) {
        return 1;
    }
    int numerator = 1;
    int denominator = 1;
    for (int i = 0; i < m; i++) {
        numerator *= (n - i);
        denominator *= (m - i);
    }
    return numerator / denominator;
}

#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;
}