三月三十日第三次实验

实验一：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <windows.h>
#define N 80

void print_text(int line, int col, char text[]); 
void print_spaces(int n); 
void print_blank_lines(int n);

int main() {
int line, col, i;
char text[N] = "hi, April~";

srand(time(0)); 

for(i = 1; i <= 10; ++i) {
line = rand() % 25;
col = rand() % 80;
print_text(line, col, text);
Sleep(1000); 
}

return 0;
}
void print_spaces(int n) {
int i;
for(i = 1; i <= n; ++i)
printf(" ");
}

void print_blank_lines(int n) {
int i;
for(i = 1; i <= n; ++i)
printf("\n");
}

void print_text(int line, int col, char text[]) {
print_blank_lines(line-1); 
print_spaces(col-1); 
printf("%s", text);
}

问题回答：该代码旨在分别随机地从1~25，1~80中取两个数作为line与col，随后打印line-1个空行后打印col-1个空格，再打印相应文本。随后程序暂停1000毫秒后继续进行。

实验二：

task2_1.c

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

int main() {
int i, n;

printf("Enter n: ");
scanf("%d", &n);

for (i = 1; i <= n; ++i)
printf("%d! = %lld\n", i, fac(i));

return 0;
}

long long fac(int n) {
static long long p = 1;
printf("p = %lld\n", p);
p = p * n;
return p;
}

实验图片：

task2_2.c

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

int main() {
int k = 4, m = 1, p1, p2;
p1 = func(k, m); 
p2 = func(k, m); 
printf("%d, %d\n", p1, p2);
return 0;
}

int func(int a, int b) {
static int m = 0, i = 2;
i += m + 1;
m = i + a + b;
return m;
}

实验图片：

问题回答：static所修饰的变量在被赋初值后会持续保留在作用域中，直到变量改变或程序结束。

实验三：

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

int main() {
int n;
long long f;
while (scanf("%d", &n) != EOF) {
f = func(n); 
printf("n = %d, f = %lld\n", n, f);
}
return 0;
}

long long func(int n)
{    
    if(n == 1)  
        return 1;
    else
        return func(n-1) * 2 + 1;        
}

实验图片：

实验四：

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

int main() {
int n, m;
while(scanf("%d%d", &n, &m) != EOF)
printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m));
return 0;
}

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

实验图片：

实验五：

迭代法：

#include <stdio.h>
double mypow(int x, int y); 
int main() {
int x, y;
double ans;
while(scanf("%d%d", &x, &y) != EOF) {
ans = mypow(x, y); 
printf("%d的%d次方: %g\n\n", x, y, ans);
}
return 0;
}

double mypow(int x, int y)
{
    double i, X = x, sum = 1;
    if(y > 0){
        for(i = 1; i <= y; i++)
            sum *= X;
        return sum;}
    else if(y < 0){
        for(i = 1; i <= -y; i++)
            sum *= 1 / X;
        return sum;}
    else if(y == 0)
        return 1;
}

实验图片：

函数递归法：

#include <stdio.h>
double mypow(int x, int y); 
int main() {
int x, y;
double ans;
while(scanf("%d%d", &x, &y) != EOF) {
ans = mypow(x, y); 
printf("%d的%d次方: %g\n\n", x, y, ans);
}
return 0;
}

double mypow(int x, int y)
{
    double X = x;
    if(y > 0)
    {
        if(y == 1)
            return x;
        else if(y != 1)
            return mypow(x, y - 1) * x;
    }
    else if(y < 0)
    {
        if(y == 1)
            return 1 / X;
        else if(y != 1)
            return mypow(X, y + 1) * 1 / X;
    }
    else
        return 1;        
}

实验图片：

实验六：

#include<stdio.h>
#include<math.h>
void hanoi(unsigned int n, char from, char temp, char to);
void moveplate(unsigned int n, char from, char to);
static double total, delta;//total定义:从程序起总共移动次数；delta定义：最近上一次移动的移动次数
int main()
{
    unsigned int n;
    int num;//定义：本次移动次数
    while(scanf("%d", &n) != EOF)
    {
    hanoi(n,'A','B','C');
    printf("\n");
    num = total - delta;
    printf("一共移动了%d次\n", num);
    delta = total;
    printf("\n");
    }
    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);total++;
}

实验图片：

实验七：

#include<stdio.h>
#include<math.h>
int is_prime(int x)
{
    int i;
    for(i = 2; i <= sqrt(x); i++)
        if(x % i == 0)
            return 0;
    return 1;
}

void print(int x)
{
    int i;
    for(i = 2; i <= x / 2; i++)
    {
        if((is_prime(i) && is_prime(x - i)))
        {
            printf("%d = %d + %d\n", x, i, x - i);
            return;
        }
    }
}

int main()
{
    int i;
    for(i = 2; i <= 20; i += 2)
        print(i);
    return 0;
}

实验图片：

实验八：

#include <stdio.h>
#include <math.h>
long func(long s);

int main() {
long s, t;
printf("Enter a number: ");
while (scanf("%ld", &s) != EOF) {
    t = func(s); 
    printf("new number is: %ld\n\n", t);
    printf("Enter a number: ");
}
return 0;
}

long func(long s)
{
    long yushu, result = 0, j = 1, i;
    for(i = 1; s != 0; i++)
    {
        yushu = s % 10;
        if(yushu % 2 == 1)
        {
            result += yushu * j;
            j *= 10;
        }
        s /= 10;
    }
    return result;
}

实验图片：