实验3

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);
    }

    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;
}

函数score_to_grade作用：计算不同分数所对应的等级

形参类型：整型  返回值类型：字符型

改变后问题：分数>=90时输出“ABCDE”;80<=分数<90时输出“BCDE”:70<=分数<80输出“CDE”;60<=分数<70输出“DE”;分数<60输出“E”

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;
}

函数功能：求各个位上数字的和

改变后的代码可以实现同样的输出

区别：改前为迭代，改后为递归

task3

#include <stdio.h>

int power(int x, int n);

int main()
{
    int n, x;
    int ans=0;

    while (printf("Enter x and n:"), scanf("%d%d", &x, &n) != EOF) {
        ans = power(x, n);
        printf("n=%d,ans=%d\n\n", x, ans);
    }
    return ans;
}

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：求一个数的n次方

此函数是递归函数，函数递归模式的数学公式为：

task4

#include<stdio.h>

int is_prime(int);

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

    for (int j = 2; j < n; ++j) {
        if (n % j == 0)
            return 0;
    }
    return 1;
}

task5

方式1：迭代

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

方式2：递归

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

task6

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

task7

#define _CRT_SECURE_NO_WARNINGS
#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) {
    int m = n*2 - 1;
    int i ,j ,s;

    for (i=1; i <= n; ++i) {

        for (s=0; s < i; ++s) {
            printf("       ");
        }
        for (j=1; j <= m; ++j) {
            printf(" O     ");
        }
        printf("\n");
        for (s=0; s < i; ++s) {
            printf("       ");
        }
        for (j=1; j <= m; ++j) {
            printf("<H>    ");
        }
        printf("\n");
        for (s=0; s < i; ++s) {
            printf("       ");
        }
        for (j=1; j <= m; ++j) {
            printf("I I    ");
        }
        m -= 2;
        printf("\n");
    }
}