实验三

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <windows.h>
#define N 80
void print_text(int line, int col, char text[]); // 函数声明
void print_spaces(int n); // 函数声明
void print_blank_lines(int n); // 函数声明
int main() {
int line, col, i;
char text[N] = "hi, November~";
srand(time(0)); // 以当前系统时间作为随机种子
for(i = 1; i <= 10; ++i) {
line = rand() % 25;
col = rand() % 80;
print_text(line, col, text);
Sleep(1000); // 暂停1000ms
}
return 0;
}
// 打印n个空格
void print_spaces(int n) {
int i;
for(i = 1; i <= n; ++i)
printf(" ");
}
// 打印n行空白行
void print_blank_lines(int n) {
int i;
for(i = 1; i <= n; ++i)
printf("\n");
}
// 在第line行第col列打印一段文本
void print_text(int line, int col, char text[]) {
print_blank_lines(line-1); // 打印(line-1)行空行
print_spaces(col-1); // 打印(col-1)列空格
printf("%s", text); 
}

在随机的行和列打印出指定字符串，共十次

#include <stdio.h>
long long fac(int n); // 函数声明
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;
p = p * n;
return p；
}

#include <stdio.h>
int func(int, int); // 函数声明
int main() {
int k = 4, m = 1, p1, p2;
p1 = func(k, m); // 函数调用
p2 = func(k, m); // 函数调用
printf("%d, %d\n", p1, p2);
return 0;
}
// 函数定义
int func(int a, int b) {
static int m = 0, i = 2;
i += m + 1;
m = i + a + b;
return m;
}

使某个变量保留每次运算后的值进行下一次运算

#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 ans;
    if(n==0)
    ans=0;
    else
    ans=func(n-1)*2+1;
    return ans;
}

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

#include<stdio.h>
int mul(int n,int m);
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    printf("%d*%d=%d\n",n,m,mul(n,m));
    return 0;
}
int mul(int n,int m){
    int ans;
    if(n==1)
    ans=m;
    else
    ans=mul(n-1,m)+m;
    return ans;
}

#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 step(int n);
int main(){
    unsigned int n;
    while(scanf("%u",&n)!=EOF){
        hanoi(n,'A','B','C');
    printf("一共移动了%d次",step(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);
    
}
int step(int n){
    if(n<=1)return 1;
    else return 2*step(n-1)+1;
    
    
}

#include <stdio.h>
int is_prime(int n);
int main()
{
    int num, f;
    int stop = 0;
    int i = 2;
    while (i < 20)
    {
        i =i+2;
        for (int j = 2; j < i; j++)
        {
            num = i - j;
            f = is_prime(num);
            if (f)
            {
                printf("%d = %d + %d\n", i, j, num);
                stop = 1;
            }
            if (stop) break;
        }
        stop = 0;
    }
 
    return 0;
}
int is_prime(int n)
{
    if (n == 1) return 0;

    for (int i = 2; i < n; i++)
    {
        int stop = 0;
        if (n % i == 0)
        {
            return 0;
            stop = 1;
        }
        if (n == 2 || (n % i != 0 && i == n - 1)) return 1;
        if (stop) break;
    }
}

#include <stdio.h>
#include<stdlib.h>
#include<math.h>
long fun(long s); // 函数声明
int main() {
long s, t;
printf("Enter a number: ");
while (scanf("%ld", &s) != EOF) {
t = fun(s); // 函数调用
printf("new number is: %ld\n\n", t);
printf("Enter a number: ");
}
return 0;
}
long fun(long s){
     int n;
    int i = 0;
    int count = 0;
    while (s!=0)
    {
        n = s % 10;
        if (n % 2 != 0)
        {
           i = n * pow(10, count) + i;
            count++;
        }
        s = s / 10;
    }
    return i;
    
}