实验三

任务1

#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); 
}

输出10次间隔1~24行、1~79空格的“hi,April~”

 

任务2

#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;
}

#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;
}

static首次调用之后变量存在，但不能被其他函数使用。再次回到函数中时使用上次结果。

任务3

 

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

 

任务4

#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)
{
static int f;
if(n>1&&m>1)
f=func(n-1,m)+func(n-1,m-1);
else
if(m>n)
f=0;
else
if(m==1)
f=n;
else
if(m==0)
f=1;
return f;
}

 

任务5

#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 a=1,i;
    if(y>0)
        for(i=y;i>0;i--)
            a=a*x;
    if(y<0)
        for(i=y;i<0;i++)
            a=a/x;
    return a;
}

任务6

#include<stdio.h>

void move(int n,int x, int y);
void hanoi(int n, char a, char b, char c);

int i=0;
int main()
{
    int n = 0;
    while(scanf("%d", &n) !=EOF )
    {
    hanoi(n, 'A', 'B', 'C');
    printf("\n一共移动了%d次\n\n",i);
    i=0;
    }
    return 0;
}

void move(int n,int x, int y)
{
    printf("%u:%c->%c\n",n,x,y);
    i++;
}

void hanoi(int n, char a, char b, char c)
{
    if (n == 1)
        move(n,a,c);
    else
    {
        hanoi(n - 1, a, c, b);
        move(n,a,c);
        hanoi(n - 1, b, a, c);
    } 
}

任务7

 

#include <stdio.h>

int is_prime(int n);

int main()
{
    int n,a=2,i;
    for(n=0;n<=20;n++)
    {
        for(;is_prime(n)==0&&n<=20;n++)
        {
            while(is_prime(a)==0||is_prime(n-a)==0)
                a++;
            printf("%d = %d + %d\n",n,a,n-a);
        }
        
    }
    return 0;
}

int is_prime(int n)
{
    int i,a=1;
    for (i=2;i<=n-1;i++) 
    { 
        if (n%i==0) 
            a=0;
    }
    return a;
}

任务8

#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 long i,a=0,t,n;
    n=1;
    for(i=0;s/n!=0;i++)
        n=10*n;    
    for(t=0;i>0;i--)
    {
        n=n/10;
        a=s/n;
        s=s-a*n;
        if(a%2==1)
        {
            t=t+a;
            t=10*t;
        }
    }
    t=t/10;
    return t;
}