实验三

task.1

点击查看代码

#include <stdio.h>

char score_to_grade(int score);


int main(){
	int score;
	int 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;
}

1.score_to_grade可将分数转化为等级，形参类型为int,返回值为char

点击查看代码

#include <stdio.h>

char score_to_grade(int score);


int main(){
	int score;
	int 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';
		case 8:     ans = 'B';
		case 7:     ans = 'C';
		case 6:     ans = 'D';
		default:    ans = 'E';
	}
	
	return ans;
}

2.改写21到28行未加break;会发生case穿透,输入什么值都会得到E

task.2

点击查看代码

#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;
}

1.计算整数各位数的和
2.能实现同样的输出，一种是迭代方式，一种是递归方式。迭代方式是循环处理，不断累加；递归方式是将问题分解计算

task.3

点击查看代码

#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;
    }
}

1.计算x的n次幂
2.是，n=0时，返回1；n为奇数时，返回xpower(x,n-1);n为偶数，返回power(x,n/2)power(x,n/2)

task.4

点击查看代码

#include <stdio.h>
int is_prime(int n) {
    if (n <= 1) {
        return 0;
    }
    if (n == 2) {
        return 1;
    }
    if (n % 2 == 0) {
        return 0;
    }
    for (int i = 3; i * i <= n; i += 2) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}
int main() {
    int count = 0; 
    printf("100以内的孪生素数：\n");
    for (int n = 2; n <= 98; n++) {
        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.\n", count);
    return 0;
}

task.5
迭代

点击查看代码

#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 sum1 = 1,sum2 = 1, sum3 = 1;
    int i, s, r;
    if (n == 0 || m == 0||m==n) return 1;
    for (i = 1; i <=n; ++i) {
        sum1 *= i;
    }
    for (s = 1; s <=m; ++s) {
        sum2 *= s;
    }
    for (r = 1; r <= n - m; ++r) {
        sum3 *= r;
    }
    return sum1 / (sum2*sum3) ;
}

递归

点击查看代码

#include <stdio.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 (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);
}

task.6

点击查看代码

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

task.7

点击查看代码

#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 = 1; i <=n; ++i) {
        for(int k=0;k<=i;k++) printf("\t");
        for (int j=2*n- 2*i + 1; j >0; j--) {
            printf("\t 0 ");
        }
        printf("\n");
        for(int k=0;k<=i;k++) printf("\t");
        for (int j=2*n- 2*i + 1; j >0; j--) {
            printf("\t<H> ");
        }
        printf("\n");
        for(int k=0;k<=i;k++) printf("\t");
        for (int j=2*n- 2*i + 1; j >0; j--) {
            printf("\tI I ");
        }
        printf("\n");
    }
}