实验3

实验任务1
task1.c
代码:
`#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;

}
`
截图:
问题1:将输入的成绩转换成等级输出。
问题2:A/B/C/D使用双引号错误,应该使用单引号,表示字符类型。case语句后无break,所有成绩都会转化成E输出。

实验任务2
task2.c
代码:
`#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;

}
`
截图:
问题1:将输入的数的各个位数的值相加。
问题2:能实现,第一个是迭代,第二个是递归。

实验任务3
task3.c
代码:
`#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;
}

}
`
截图:

问题1:输出x的n次方。
问题2:是递归函数。

实验任务4
task4.c
代码:
`#include <stdio.h>

include<math.h>

int is_prime(int n);
int main() {
int sum=0;
for (int n = 1; n <= 100; n++) {
if (is_prime(n) &&_prime(n + 2)) {
sum += 1;
printf("100以内的孪生素数:\n%d %d\n",n,n+2);

	}
	
}
printf("100以内的素数共有%d个\n",sum);
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;

`
截图:

实验任务5
task5.c
代码:
`#include <stdio.h>

include <stdlib.h>

void hanoi(unsigned int n, char from, char temp, char to);
void moveplate(unsigned int n, char from, char to);
int i = 0;
int main()
{
unsigned int n;
while (scanf("%u", &n) != EOF)
{
i = 0;
hanoi(n, 'A', 'B', 'C');
printf("\n一个移动了:%d次\n", i);
printf("\n");
}
system("pause");
return 0;
}
void hanoi(unsigned int n, char from, char temp, char to)
{
if (n == 1)
moveplate(n, from, to);
else
{
hanoi(n - 1, from, to, temp);
moveplate(n, from, to);
hanoi(n - 1, temp, from, to);
}
}
void moveplate(unsigned int n, char from, char to)
{
printf("%u:%c-->%c\n", n, from, to);
i++;
}
`
截图:

实验任务6
task6.c
代码(迭代):
`#include <stdio.h>

include <stdlib.h>

int func(int n, int m); // 函数声明

int main() {
int n, m;
int ans;

while (scanf("%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 i;
int ans1 = 1, ans2 = 1, ans3;
for (i = n; i > n - m; --i) {
ans1 = ans1 * i;
}
for (i = m; i > 0; --i) {
ans2 = ans2 * i;
}
ans3 = ans1 / ans2;
return ans3;
}
代码(递归):#include <stdio.h>

include <stdlib.h>

int func(int n, int m); // 函数声明

int main() {
int n, m;
int ans;

while (scanf("%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 (n < m) {
return 0;
}
if (m == 0 || m == n) {
return 1;
}

return func(n - 1, m) + func(n - 1, m - 1);

}
`
截图:

实验任务7
task7.c
代码:
`#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 i = a;
if (i > b) {
i = b;
if (i > c) {
i = c;
}
}

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

}

`
截图:

posted @ 2025-04-04 15:52  汪天昊  阅读(22)  评论(0)    收藏  举报