实验任务3

实验1:

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#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;
}

捕获
Q1:函数功能是实现分数到等级到转化;形参是int整数类型,返回值是char字符型。
Q2:如果去掉了break,那么无论输入什么数字,代码都是读取到最下面到代码,即结果是E,程序错误。

实验2:

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#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;
}

1
Q1:作用是实现整数的各数字之和。
Q2:能,运用的是递归算法,上面的是迭代算法。

实验3:

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#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;
	}
}

2
Q1:power功能是实现数的次方运算。(如2的3次方,3的5次方)
Q2:
9E2D2F50A53657861C23C9F4CA860F4B

实验4:

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#include <stdio.h>
int is_prime (int a);

int main ()
{
	int num,max = 100;
	int sum = 0;
	int n1,n2;
	printf("100以内的孪生素数:");
	for(num = 1;num + 2 <= max;num ++)
	{
		if(is_prime(num) == 1 && is_prime(num + 2) == 1)
	    {
	    	n1 = num;
	    	n2 = num + 2;
		    printf("%d %d\n",n1,n2);
		    sum ++;
	    }
	}
    printf("100以内的孪生素数共有%d个\n",sum);
	
	return 0;
}

int is_prime (int a)
{
	int i;
	if(a == 1)
	return 0;
	else if(a == 2)
	return 1;
	else 
	{
		if(a % 2 == 0)
		return 0;
		else
		{
			for(i = 3; i*i <= a; i+=2)
			{
				if(a % i == 0)
				{
					return 0;
					break;
				}
			}
			return 1;
		}
	}

}


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实验5:
(1)迭代算法:

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#include <stdio.h>
int func(int n, int m);
/*迭代*/ 
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 up = 1, down = 1;
	int i,k;
	int c;
	for(k = n-m+1;k<=n;k++)
	{
		up = k*up;
	}
	for(i = 1;i<=m;i++)
	{
		down = i *down;
	}
	c = up/down;
	return c;
} 

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(2)递归算法:

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

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实验6:

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#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;
	if(a < b && a < c)
	i = a;
	else if (b < a && b < c)
	i = b;
	else
	i = c;
	
	for(;i >= 1;i--)
	{
		if(i == 1)
		{
			return i;
			break;
		}
		if (a % i == 0 && b % i == 0 && c % i == 0)
		return i;
	}
	
}

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实验7:

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#include <stdio.h>
#include <stdlib.h>
void print_charman(int n); 

int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); 
       
    return 0;
}

void print_charman(int n)
{ 
	for (int i = 0;i < n; i++)
	{
		int line;
		int a;
		for(line = 0;line < 2 * i; line++)
	    {
	    	printf("   ");
		}
		int number1,number2,number3;
		number1 = 2 * (n - i) - 1;
		number2 = number1;
		number3 = number2;
	    for(a = 0;a < number1; a++)
	    { 
	    	printf(" o ");
	    	if (a != number1)
	    	printf("   ");
		}
		printf("\n");
		for(line = 0;line < 2 * i; line++)
	    {
	    	printf("   ");
		}
		for(a = 0;a < number2 ;a++)
		{
			printf("<H>");
			if (a != number2)
	    	printf("   ");
		}
		printf("\n");
		for(line = 0;line < 2 * i; line++)
	    {
	    	printf("   ");
		}
		for(a = 0;a < number3; a++)
		{
		    printf("I I");
		    if (a != number3)
	    	printf("   ");
	    }
	    printf("\n");
	}
}

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posted @ 2025-10-31 17:04  hhqdream  阅读(1)  评论(0)    收藏  举报