实验三
实验任务1
实验代码
#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);
}
实验结论
回答问题
1：每隔一秒在输出窗口的随机位置生成一个“hi, April~”，共生成十个
实验任务2
task2.1实验代码
#include <stdio.h>
long long fac(int n);
int main()
{
int i,n;
printf("Enyer 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;
}
task2.1实验结论
task2.2实验代码
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;
}
实验结论
回答问题
static变量在函数第一次被调用时第一次遇到该变量时进行初始化，这也是唯一的一次初始化，此后该变量的初始化语句不会再被执行。下一次调用时使用上次的保存值。
实验任务3
实验代码
#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;
if(n==2)
return 3;
return 2*func(n-1)+1;
}
实验结论
回答问题
实验任务4
实验代码
#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 == 0 || n == m || m == 0)
return 1;
else if (n < m)
return 0;
else
return func(n - 1, m) + func(n - 1, m - 1);
}
实验结论
实验任务5
task5.1实验代码
#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 result = 1.0;
int i;
if(y >= 0) {
for(i = 0; i < y; i++) {
result *= (double)x;
}
}
else {
for(i = 0; i < -y; i++) {
result /= (double)x;
}
}
return result;
}
实验结论
task5.2实验代码
#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){
if(y>=0){
if(y==0)
return 1;
return mypow(x,y-1) * x;
}
if(y<0){
double z = 1.0/x;
return mypow(x,y+1) * z;
}
}
实验结论
实验任务6
实验代码
#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 total=0;
int main() {
unsigned int n;
while (scanf("%u", &n) != EOF){
hanoi(n, 'A', 'B', 'C');
printf("一共移动了%d次",total);
total = 0;
}
system("pause");
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++;
}
实验结论
实验任务7
实验代码
#include <stdio.h>
int is_prime(int n);
int main() {
int num = 4;
while (num <= 20) {
printf("%d = ", num);
int num1 = 2;
int num2 = num - num1;
while (num1 <= num2) {
if (is_prime(num1) && is_prime(num2)) {
printf("%d + %d", num1, num2);
break;
}
num1++;
num2--;
}
num += 2;
printf("\n");
}
return 0;
}
int is_prime(int n) {
if (n <= 1) {
return 0;
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return 0;
}
}
return 1;
}
实验结论
实验任务8
实验代码
#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 newone= 0;
long weishu = 1;
while (s > 0) {
int oldone = s % 10;
if (oldone % 2 != 0) {
newone += oldone * weishu;
weishu *= 10;
}
s /= 10;
}
return newone;
}
实验结论
