实验3

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,november~";
srand(time(0));
for(i=0;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);}

2

 

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

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

#include<stdio.h>
#include<stdlib.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);
    
    system("pause");
    return 0;
}
int func(int a,int b){
    static int m=0,i=2;
    i+=m+1;
    m=i+a+b;
    return m;}

#include<stdio.h>
long long func(int n);
  
int main() {
    int n;
    long long f;
    while (scanf_s("%d", &n) != EOF)
    {
        f = func(n);
        printf("n=%d,f=%11d\n", n, f);
    }
    return 0;
}

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

}

#include <stdio.h>
int func(int n, int m);
int main() {
    int n, m;
    while (scanf_s("%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) {
    int l;
    if (n < m)
        return 0;
    else if (n == m || m == 0)
         return 1;
      else
      return l = func(n - 1, m) + func(n - 1, m - 1);
}

#include<stdio.h>
int mul(int n, int m);
int main() {
    int m, n;
    while (scanf_s("%d%d", &n, &m) != EOF)
        printf("%d*%d=%d\n", n, m, mul(n, m));
        return 0;
}

int mul(int n, int m) {
    if (m == 0 || n == 0)
        return 0;
    int l=n;
    if (m == 1)
        return n;
    else
        return l += mul(n, m - 1);
}

#include<stdio.h>
int a = 0;
void hanoi(int i, char one, char two, char three);
int main() {
    int n;

    while (scanf_s("%d", &n) != EOF)
    {
        hanoi(n, 'A', 'B', 'C');

        printf("一共移动%d次\n", a);
    }
    return 0;
}
void hanoi(int i, char one, char two, char three) {
    void move(int x, int y);

    if (i == 1) {
        move(one, three);
    }
    else {
        hanoi(i - 1, one, three, two);
        move(one, three);
        hanoi(i - 1, two, one, three);
    }
}
void move(int x, int y) {
    a++;
    printf("%c-->%c\n", x, y);

}

#include<stdio.h>
#include<math.h>
int is_prime(int n);
int main() {
    int n, m, s;
    while (scanf_s("%d", &s) != EOF)
    {
        
            for (m = 2; m < 20; m++)
            {
                for (n = 2; n < 20; n++)
                {
                    if (is_prime(m) * m + is_prime(n) * n == s)
                        break;
                }
                if (is_prime(m) * m + is_prime(n) * n == s)
                    break;
            }

        printf("%d=%d+%d\n", s, m, n);
    }
    return 0;
}

int is_prime(int n) {
    int i;
    for (i = 2; i <= sqrt(n); i++)
    {
        if (n % i == 0)
            return 0;
    }
    if (n == 2)
    return 1;
    if (i > sqrt(n))
        return 1;
}

#include <stdio.h>
#include<math.h>
long func(long s);
int main() {
    long s, t;
    printf("Enter a number: ");
    while (scanf_s("%ld", &s) != EOF) {
        t = func(s); 
        printf("new number is: %ld\n\n", t);
        printf("Enter a number: ");
    }
    return 0;
}

long func(long s) {
    int x[50],i=0,l,m=0,n;
    while (s != 0)
    {
        l = s % 10;
        if (l % 2 == 1)
            x[i++] = l;
     s = s / 10;
    }
    for (n = 0; n < i; n++)
        m += x[n]* pow(10, n);
        return m;
}