task1
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <time.h>
4 #include <windows.h>
5 #define N 80
6 void print_text(int line, int col, char text[]); // 函数声明
7 void print_spaces(int n); // 函数声明
8 void print_blank_lines(int n); // 函数声明
9 int main() {
10 int line, col, i;
11 char text[N] = "hi, November~";
12 srand(time(0)); // 以当前系统时间作为随机种子
13 for (i = 1; i <= 10; ++i) {
14 line = rand() % 25;
15 col = rand() % 80;
16 print_text(line, col, text);
17 Sleep(1000); // 暂停1000ms
18 }
19 return 0;
20 }
21 // 打印n个空格
22 void print_spaces(int n) {
23 int i;
24 for (i = 1; i <= n; ++i)
25 printf(" ");
26 }
27 // 打印n行空白行
28 void print_blank_lines(int n) {
29 int i;
30 for (i = 1; i <= n; ++i)
31 printf("\n");
32 }
33 // 在第line行第col列打印一段文本
34 void print_text(int line, int col, char text[]) {
35 print_blank_lines(line - 1); // 打印(line-1)行空行
36 print_spaces(col - 1); // 打印(col-1)列空格
37 printf("%s", text); // 在第line行、col列输出text中字符串
38 }
如下图生成（还有4个未显示）
代码功能：随机在某行某列打印“hi，November”，并在其之后的某行与某列再次打印，重复9次，共打印10个
task2.1
1 #include <stdio.h>
2 long long fac(int n); // 函数声明
3 int main() {
4 int i, n;
5 printf("Enter n: ");
6 scanf("%d", &n);
7 for (i = 1; i <= n; ++i)
8 printf("%d! = %lld\n", i, fac(i));
9 return 0;
10 }
11 // 函数定义
12 long long fac(int n) {
13 static long long p = 1;
14 //printf("p = %lld\n", p);
15 p = p * n;
16 return p;
17 }
task2.2
1 #include <stdio.h>
2 int func(int, int); // 函数声明
3 int main() {
4 int k = 4, m = 1, p1, p2;
5 p1 = func(k, m); // 函数调用
6
7 p2 = func(k, m); // 函数调用
8 printf("%d, %d\n", p1, p2);
9 return 0;
10 }
11 // 函数定义
12 int func(int a, int b) {
13 static int m = 0, i = 2;
14 i += m + 1;
15 m = i + a + b;
16 return m;
17 }
1.首先在p1中，i=2+0+1=3，m=i+a+b=3+1+4=8
2.之后函数外面的m还是1，k还是4，但是函数里面的m是8，i是3
3.p2中，i=i+m+1=3+8+1=12，m=i+a+b=12+1+4=17
总结：局部static变量可在整个代码中循环，但是作用域只局限在局部
task3
1 #include <stdio.h>
2 long long func(int n); // 函数声明
3 int main() {
4 int n;
5 long long f;
6 while (scanf("%d", &n) != EOF) {
7 f = func(n); // 函数调用
8 printf("n = %d, f = %lld\n", n, f);
9 }
10 return 0;
11 }
12
13 long long func(int n) {
14 long long sum = 1;
15 if (n == 1) {
16 sum = 1;
17 }
18 else if (n > 1) {
19 sum = 2 * func(n - 1) + 1;
20 }
21 return sum;
22 }
task4
1.迭代：
1 #include <stdio.h>
2 int func(int n, int m);
3 int main() {
4 int n, m;
5 while (scanf("%d%d", &n, &m) != EOF)
6 printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m));
7 return 0;
8 }
9
10 int func(int n, int m) {
11 int sum1 = 1;
12 int sum2 = 1;
13 int sum3 = 1;
14
15 int s = n - m;
16 if (s < 0) {
17 return 0;
18 }
19
20 for (n; n > 0; n--) {
21 sum1 *= n;
22 }
23
24 for (m; m > 0; m--) {
25 sum2 *= m;
26 }
27
28 for (s; s > 0; s--) {
29 sum3 *= s;
30 }
31
32 int sum = sum1 / (sum2 * sum3);
33
34 return sum;
35
36 }
2.递归
1 #include <stdio.h>
2 int func(int n, int m);
3 int main() {
4 int n, m;
5 while (scanf("%d%d", &n, &m) != EOF)
6 printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m));
7 return 0;
8 }
9
10 int func(int n, int m) {
11 if (n < m) {
12 return 0;
13 }
14 int sum;
15 if (m == 0) {
16 return 1;
17 }
18 if (n == 1 && m == 1) {
19 sum = 1;
20 }
21 else if (m == 1) {
22 sum = n;
23 }
24 else {
25 sum = func(n - 1, m) + func(n - 1, m - 1);
26 }
27
28 return sum;
29 }
两者结果一样：
task5
1 #include <stdio.h>
2
3 void move(int n, char A, char B, char C);
4
5 int main()
6 {
7 int n;
8 char A, B, C;
9 while (scanf("%d", &n) != EOF) {
10 int step = 0;
11 move(n, 'A', 'B', 'C');
12 for (n; n > 0; n--) {
13 step = 2 * step + 1;
14 }
15 printf("\n");
16 printf("一共移动了%d次\n", step);
17 printf("\n");
18 }
19
20 return 0;
21 }
22
23 void move(int n, char A, char B, char C) {
24 if (n == 1) {
25 printf("%d：%c --> %c\n", n, A, C);
26 }
27 else {
28 move(n - 1, 'A', 'C', 'B');
29 printf("%d：%c --> %c\n", n, A, C);
30 move(n - 1, 'B', 'A', 'C');
31 }
32 }
tip：
三柱汉诺塔移动n个盘的步数是（n-1）盘的步数的2倍多1；
step(n)=2*step(n-1)+1；
task6
1 #include <stdio.h>
2 #include <math.h>
3 long func(long s); // 函数声明
4 int main() {
5 long s, t;
6 printf("Enter a number: ");
7 while (scanf("%ld", &s) != EOF) {
8 t = func(s); // 函数调用
9 t = func(t);
10 printf("new number is: %ld\n\n", t);
11 printf("Enter a number: ");
12 }
13 return 0;
14 }
15
16 long func(long s) {
17 int n;
18 long sum = 0;
19 while (s > 0) {
20 n = s % 10;
21 if (n % 2 != 0) {
22 sum = sum * 10 + n;
23 }
24 s /= 10;
25 }
26
27 return sum;
28 }
task7
1 #include <stdio.h>
2 int func(int n, int m);
3 int main()
4 {
5 int n = 1;
6 int flag = 0;
7 while (flag == 0) {
8 int n2 = n * n;
9 int n3 = n * n * n;
10 flag = func(n2, n3);
11 if (flag == 1) {
12 printf("result=%d\n", n);
13 printf("它的平方是%d*%d=%d\n", n, n, n2);
14 printf("它的立方是%d*%d*%d=%d\n", n, n, n, n3);
15 return 0;
16 }
17 else {
18 n++;
19 continue;
20 }
21 }
22 return 0;
23 }
24
25 int func(int n, int m) {
26 int k = 0;
27 int s;
28 int jugde[10];
29 for (s = 0; s < 10; s++) {
30 jugde[s] = 1;
31 }
32 while (n > 0) {
33 s = n % 10;
34 switch (s) {
35 case 0:
36 jugde[s]--;
37 break;
38 case 1:
39 jugde[s]--;
40 break;
41 case 2:
42 jugde[s]--;
43 break;
44 case 3:
45 jugde[s]--;
46 break;
47 case 4:
48 jugde[s]--;
49 break;
50 case 5:
51 jugde[s]--;
52 break;
53 case 6:
54 jugde[s]--;
55 break;
56 case 7:
57 jugde[s]--;
58 break;
59 case 8:
60 jugde[s]--;
61 break;
62 case 9:
63 jugde[s]--;
64 break;
65 }
66 n /= 10;
67 }
68 while (m > 0) {
69 s = m % 10;
70 switch (s) {
71 case 0:
72 jugde[s]--;
73 break;
74 case 1:
75 jugde[s]--;
76 break;
77 case 2:
78 jugde[s]--;
79 break;
80 case 3:
81 jugde[s]--;
82 break;
83 case 4:
84 jugde[s]--;
85 break;
86 case 5:
87 jugde[s]--;
88 break;
89 case 6:
90 jugde[s]--;
91 break;
92 case 7:
93 jugde[s]--;
94 break;
95 case 8:
96 jugde[s]--;
97 break;
98 case 9:
99 jugde[s]--;
100 break;
101 }
102 m /= 10;
103 }
104 for (s = 0; s < 10; s++) {
105 if (jugde[s] != 0) {
106 k = 0;
107 break;
108 }
109 else {
110 k = 1;
111 }
112 }
113 return k;
114 }
标签：case,jugde,int,break,--,实验,func From： https://www.cnblogs.com/VitaminC114514/p/17798113.html