实验3

task1

#include <stdio.h>
#include <stdlib.h>
char score_to_grade(int score);  // 函数声明

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);
    }
    system("pause");
    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;
}

 

功能是将score转化为对应的等级

形参类型为int 返回值类型为char

错误：字符串应该用单引号 缺少break语句 

task2

#include <stdio.h>

int sum_digits(int n);  // 函数声明

int main() {
    int n;
    int ans;

    while(printf("Enter n: "), scanf("%d", &n) != EOF) {
        ans = sum_digits(n);    // 函数调用
        printf("n = %d, ans = %d\n\n", n, ans);
    }

    return 0;
}

// 函数定义
int sum_digits(int n) {
    int ans = 0;

    while(n != 0) {
        ans += n % 10;
        n /= 10;
    }

    return ans;
}

 

功能是计算一个整数n的各位数字之和

能 一个是递归，一个是迭代

task3

#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;
    }
}

task4

#include <stdio.h>
#include <stdlib.h>

int is_prime(int n);

int main()
{
    int count;
    int n;
    n = 2;
    count = 0;
    printf("100以内的孪生素数：\n");
    while(n < 99)
    {
        if(is_prime(n) && is_prime(n + 2))
        {
            printf("%d  %d\n", n, n+2);
            count++;
        }
        n = n + 1;
    }
    printf("100以内的孪生素数共有%d个\n",count);
    system("pause");
    return 0;
}

int is_prime(int n){
    int i;
    i = 2;
    while(i < n)
    {
        if(n % i == 0)
            break;
        else
            i = i + 1;
    }
    if(i == n)
        return 1;
    else
        return 0;

}

 

task5

#include <stdio.h>
#include<stdlib.h>
void hanoi(unsigned int n,char from,char temp,char to,unsigned int *count);
void moveplate(unsigned int n,char from,char to,unsigned int *count);
int main()
{
    unsigned int n;
    unsigned int count;
    while(1){
        scanf("%u",&n);
        count = 0;
        hanoi(n,'A','B','C',&count);
        printf("一共移动了%u次\n\n", count);
    }
    system("pause");
    return 0;
}
void hanoi(unsigned int n,char from,char temp,char to,unsigned int *count)
{
    if(n == 1)
        moveplate(n,from,to,count);
    else
    {
        hanoi(n-1,from,to,temp,count);
        moveplate(n,from,to,count);
        hanoi(n-1,temp,from,to,count);
    }
}
void moveplate(unsigned int n,char from,char to,unsigned int *count)
{
    (*count)++;
    printf("%u:%c-->%c\n",*count,from,to);
}

 

task6

#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);
    }
    system("pause");
    return 0;
}
int func(int n, int m){
    int i;
    long long res;
    if(m < 0 || m > n)
        return 0;
    if(m == 0 || m == n)
        return 1;
    res = 1;
    for (i = 1;i <= m;i++){
        res = res * (n - m + i) / i;
    }
    return (int)res;
}

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

 

task7

#include <stdio.h>
#include <stdlib.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);
    }
    system("pause");
    return 0;
}


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