实验3

1

#include<stdio.h>

char score_to_grade(int score);

int main()
{
    int score;
    char grade;

    while (scanf_s("%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）将输入的分数对应一个等级；形参类型是一个整数类型的分数；返回值是一个字符类型。

 （2）ans的类型和赋值不匹配；没有break会继续执行case；defualt中是'E',前面是"A"。

2

#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;
    }
}

 （1）计算输入数中各数字的和。

 （2）效果一样，原来的是运用了迭代，改后是递归。

3

#include <stdio.h>

int power(int x, int n);
int main() {
    int x, n;
    int ans;

    while (printf("Enter x and n: "), scanf_s("%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;
    }
}

 （1）用来计算x^n

 （2）是递归，

 

4

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

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

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

 

5

#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 main()
{
    unsigned int n,i;
    int count = 0;
    int j=1;
    while (scanf_s("%u", &n) != EOF)
    {
        hanoi(n, 'A', 'B', 'C');
        for (i = 0; i < n; ++i)
        {
            j *= 2;
        }
        count = j - 1;
        printf(" \n");
        printf("一共移动了%d次.", count);
        count = 0;
    }
    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);
}

 

6（方法1、方法2）

#include <stdio.h>
int func(int n, int m);  

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

    while (scanf_s("%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=n, down=m;
    int c=0;
    if (m == 0 || m == n)
        return 1;
    while ((m > 1)and (n > m - 1))
    {
        --n, --m;
        up *= n;
        down *= m;
    }
    if(down>0)
        c = up / down;
    return c;
}

#include <stdio.h>
int func(int n, int m);

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

    while (scanf_s("%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 c1, c2;
    int c=0;
    int up1 = n - 1 ,down1 = m;
    int up2 = n - 1, down2 = m-1;
    if (n < m)
        return 0;
    if (m == n or m == 0)
        return 1;
    if (m == 1)
        return n;
    else
        return func(n - 1, m) + func(n - 1, m - 1);
    while ( (n > m - 1) and (m > 1))
    {
        --n, --m;
        up1 *= n,up2 *=n;
        down1 *= m,down2*=m;
    }
    c1 = up1 / down1;
    c2 = up2 / down2;
    c = c1 + c2;
    return c;
}

 

7

#include <stdio.h>

int gcd(int a, int b, int c);


int main() {
    int a, b, c;
    int ans;

    while (scanf_s("%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) {
    while (a != b)
    {
        if (a > b)
            a = a - b;
        else
            b = b - a;
    }
    while (c != b)
    {
        if (c > b)
            c = c - b;
        else
            b = b - c;
    }
    while (a != c)
    {
        if (a > c)
            a = a - c;
        else
            c = c - a;
    }
    return a;
}