1.实验任务1
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<time.h>
4 #include<windows.h>
5 #define N 80
6
7 void print_text(int line, int col, char text[]);
8 void print_spaces(int n);
9 void print_blank_lines(int n);
10
11 int main()
12 {
13 int line,col,i;
14 char text[N] = "hi, April~";
15
16 srand(time(0));
17
18 for(i = 1;i <= 10;++i)
19 {
20 line = rand() % 25;
21 col = rand() % 80;
22 print_text(line,col,text);
23 Sleep(1000);
24 }
25
26 return 0;
27
28 }
29
30
31 void print_spaces(int n)
32 {
33 int i;
34
35 for(i = 1;i<=n; ++i)
36 printf(" ");
37 }
38
39
40 void print_blank_lines(int n)
41 {
42 int i;
43
44 for(i = 1;i <= n;++i)
45 printf("\n");
46 }
47
48
49 void print_text(int line, int col, char text[])
50 {
51 print_blank_lines(line-1);
52 print_spaces(col-1);
53 printf("%s",text);
54 }
在屏幕上二维坐标点随机生成“hi,April~”，单个语句长度为8个字符，N为80，总共生成十个语句。
2.实验任务2
task2_1.c
1 #include<stdio.h>
2 #include<stdlib.h>
3 long long fac(int n);
4
5 int main()
6 {
7 int i,n;
8
9 printf("Enter n: ");
10 scanf("%d",&n);
11 for(i = 1;i<=n; ++i)
12 printf("%d! = %lld\n",i,fac(i));
13
14 system("pause");
15 return 0;
16 }
17
18
19 long long fac(int n)
20 {
21 static long long p = 1;
22
23 p = p * n;
24
25 return p;
26 }
task2_2.c
1 #include<stdio.h>
2 #include<stdlib.h>
3
4 int func(int, int );
5
6 int main()
7 {
8 int k = 4,m = 1,p1,p2;
9
10 p1 = func(k,m);
11 p2 = func(k,m);
12 printf("%d, %d\n",p1,p2);
13
14 system("pause");
15 return 0;
16 }
17
18
19 int func(int a, int b)
20 {
21 static int m = 0,i =2;
22
23 i = i + m + 1;
24 m = i + a + b;
25
26 return m;
27 }
局部static变量的特性：数据存储在静态存储区，函数退出时变量始终存在，再次进入该函数时，将不再重新赋值，直接使用上一次的结果。
3.实验任务3
1 #include<stdio.h>
2 #include<stdlib.h>
3
4 long long func(int n);
5
6 int main()
7 {
8 int n;
9 long long f;
10
11 while(scanf("%d",&n)!=EOF)
12 {
13 f = func(n);
14 printf("n = %d,f = %lld\n",n,f);
15 }
16
17 system("pause");
18 return 0;
19 }
20
21
22 long long func(int n)
23 {
24 int i=1;
25 long long p=1;
26 for(;i<=n;i++)
27 {
28 p = p * 2;
29
30 }
31 return p-1;
32 }
4.实验任务4
1 #include<stdio.h>
2 #include<stdlib.h>
3 int func(int n,int m);
4
5 int main()
6 {
7 int n,m;
8
9 while(scanf("%d%d",&n,&m)!=EOF)
10 printf("n = %d,m = %d,ans = %d\n",n,m,func(n,m));
11
12 system("pause");
13 return 0;
14 }
15
16 int func(int n,int m)
17 { int s;
18 if(n==m||n!=0&&m==0)
19 return 1;
20 if(n<m)
21 return 0;
22 else
23 {s=func(n-1,m)+func(n-1,m-1);
24 return s;}
25 }
5.实验任务5
task5_1.c
1 #include<stdio.h>
2 #include<stdlib.h>
3 double mypow(int x,int y);
4
5 int main()
6 {
7 int x,y;
8 double ans;
9
10 while(scanf("%d%d",&x,&y)!=EOF)
11 {
12 ans = mypow(x,y);
13 printf("%d的%d次方:%g\n\n",x,y,ans);
14 }
15
16 system("pause");
17 return 0;
18 }
19
20 double mypow(int x,int y)
21 {
22 double s = 1;
23 int i = 1;
24 if(y>=0)
25 {for(;i<=y;i++)
26 {s = s * x;}
27 }
28 if(y<0)
29 {for(;i<=abs(y);i++)
30 {s = s / x;}
31 }
32
33 return s;
34 }
task5_2.c
1 #include<stdio.h>
2 #include<stdlib.h>
3 double mypow(int x,int y);
4
5 int main()
6 {
7 int x,y;
8 double ans;
9
10 while(scanf("%d%d",&x,&y)!=EOF)
11 {
12 ans = mypow(x,y);
13 printf("%d的%d次方:%g\n\n",x,y,ans);
14 }
15
16 system("pause");
17 return 0;
18 }
19
20 double mypow(int x,int y)
21 {
22 double s=1;
23 if(y==0)
24 return 1;
25 if(y>0)
26 {s = x * mypow(x,y-1);
27 return s;}
28 if(y<0)
29 {s = s / mypow(x,-y);
30 return s;}
31 }
6.实验任务6
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<math.h>
4 void hanoi(unsigned int n,char from,char temp,char to);
5 void moveplate(unsigned int n,char from,char to);
6 int main()
7 {
8 unsigned int n;
9 int m;
10 while(scanf("%u",&n)!=EOF)
11
12 {
13 m = pow(2,n);
14 hanoi(n,'A','B','C');
15 printf("一共移动了%d次\n\n",m-1);
16 }
17
18 system("pause");
19 return 0;
20 }
21 void hanoi(unsigned int n,char from,char temp,char to)
22 {
23 if(n==1)
24 moveplate(n,from,to);
25 else
26 {
27 hanoi(n-1,from,to,temp);
28 moveplate(n,from,to);
29 hanoi(n-1,temp,from,to);
30 }
31 }
32 void moveplate(unsigned int n,char from,char to)
33 {
34 printf("%u:%c-->%c\n",n,from,to);
35 }
7.实验任务7
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<math.h>
4 int is_prime(int n);
5 int main()
6 {
7 int n,i,p = 1,q,p_flag = 1,q_flag = 1;
8 for(n=2;n<=20;n++)
9 { if(n%2 == 1)
10 continue;
11 if(n>2&&n%2 == 0)
12 { p = 1;
13 do
14 {
15 p = p + 1;
16 q = n - p;
17 p_flag = is_prime(p);
18 q_flag = is_prime(q);
19 }while(p_flag * q_flag == 0);
20 printf("%d = %d + %d\n",n,p,q);
21 }
22 }
23 system("pause");
24 return 0;
25 }
26
27 int is_prime(int n)
28 {
29 int i;
30 for(i=2;i<=sqrt(n);++i)
31 {
32 if(n%i==0)
33 break;
34 }
35 if(n>1&&i>sqrt(n))
36 return 1;
37 else
38 return 0;
39
40 }
8.实验任务8
1 #include<stdio.h>
2 #include<stdlib.h>
3 long long func(long s);
4 int main()
5 {
6
7 long s,t;
8
9 printf("Enter a numner:");
10 while(scanf("%ld",&s)!=EOF)
11 {
12 t = func(s);
13 printf("new number is: %ld\n\n",t);
14 printf("Enter a number:");
15 }
16
17 system("pause");
18 return 0;
19 }
20
21 long long func(long s)
22 {
23 int d;
24 long s1 = 1;
25 long m = 0;
26 while(s>0)
27 {
28 d = s % 10;
29 if(d%2)
30 {
31 m = s1 * d + m;
32 s1 = s1 * 10;
33 }
34 s = s / 10;
35 }
36 return m;
37 }
标签：return,int,long,实验,func,printf,include From： https://www.cnblogs.com/cxj114-514/p/17280250.html