实验3
一、实验任务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[]); ?// 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); ? ? ? ? // 在第line行、col列输出text中字符串
}

该程序的目的是在命令行窗口中以随机位置输出字符串"hi, November~",每次输出后暂停1秒。
二、实验任务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); // 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;
}

与我想的一致
static关键字的作用包括限定变量或函数的作用范围、保留变量的值和扩展变量的生命周期。
三、实验任务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)
{
if(n==0)
return 0;
else
return (func(n-1)+1)*2-1;
}

四、实验任务4
1、迭代
#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)
return 0;
int num= 1;
int deno= 1;
for (int i = 0; i < m; i++)
num *= n - i;
for (int i = 1; i <= m; i++)
deno*= i;
return num/ deno;
}

2、递归
#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)
return 0;
if (m == 0 || n == m)
return 1;
return func(n - 1, m - 1) + func(n - 1, m);
}

五、实验任务5
#include <stdio.h>
#include <stdlib.h>
int hanoi(int n, char x, char y, char z, int *counter)
{
int move(char, int, char);
if (n == 1)
{
move(x, 1, z);
(*counter)++;
}
else
{
hanoi(n - 1, x, z, y, counter);
move(x, n, z);
(*counter)++;
hanoi(n - 1, y, x, z, counter);
}
return 0;
}
int move(char get, int n, char put)
{
static int k = 1;
printf("%d: %c --> %c\n", n, get, put);
k++;
return 0;
}
int main()
{
system("color CF");
int hanoi(int, char, char, char, int *);
int n, counter;
while (scanf("%d", &n) != EOF)
{
counter = 0;
hanoi(n, 'A', 'B', 'C', &counter);
printf("\n一共移动了%d次\n\n", counter);
}
return 0;
}

六、实验任务6
#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); // 函数调用
printf("new number is: %ld\n\n", t);
printf("Enter a number: ");
}
return 0;
}
long func(long s) {
long result = 0;
long multiplier = 1;
while (s > 0) {
long digit = s % 10;
if (digit % 2 != 0) {
result = result + digit * multiplier;
multiplier *= 10;
}
s /= 10;
}
return result;
}

六、实验任务6
#include <stdio.h>
#include <stdlib.h>
int fuck(long n ,long m);
int main(void)
{
for (int i=0;;i++){
int a=i*i;
int b=i*i*i;
if(fuck(a,b)==1){
printf("%d",i);
break;
}
}
return 0;
}
int fuck(long n,long m){
long temp = n;
int Di= 0;
while (temp > 0) {
temp /= 10;
Di++;
}
int a[Di];
for (int i = Di - 1; i >= 0; i--) {
a[i] = n% 10;
n/= 10;
}
long temp2 = m;
int Di2= 0;
while (temp2 > 0) {
temp2 /= 10;
Di2++;
}
int b[Di2];
for (int i = Di2 - 1; i >= 0; i--) {
b[i] = m% 10;
m/= 10;
}
int v= Di+Di2;
int c[v];
int i, j, k;
i = j = k = 0;
while (i < Di){
c[k++] = a[i++];
}
while (j<Di2){
c[k++] = b[j++];
}
int cao[10]={0};
for(int i=0;i<v;i++){
cao[c[i]]++;
}
for(int i=0;i<=9;i++)
{
if(cao[i]!=1)
return 0;
}
return 1;
}


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