This commit is contained in:
krahets
2023-09-10 18:37:26 +08:00
parent 4307372a5b
commit e48716c883
6 changed files with 192 additions and 19 deletions

View File

@@ -158,7 +158,15 @@ status: new
=== "C"
```c title="iteration.c"
[class]{}-[func]{forLoop}
/* for 循环 */
int forLoop(int n) {
int res = 0;
// 循环求和 1, 2, ..., n-1, n
for (int i = 1; i <= n; ++i) {
res += i;
}
return res;
}
```
=== "Zig"
@@ -344,7 +352,17 @@ status: new
=== "C"
```c title="iteration.c"
[class]{}-[func]{whileLoop}
/* while 循环 */
int whileLoop(int n) {
int res = 0;
int i = 1; // 初始化条件变量
// 循环求和 1, 2, ..., n-1, n
while (i <= n) {
res += i;
i++; // 更新条件变量
}
return res;
}
```
=== "Zig"
@@ -539,7 +557,19 @@ status: new
=== "C"
```c title="iteration.c"
[class]{}-[func]{whileLoopII}
/* while 循环(两次更新) */
int whileLoopII(int n) {
int res = 0;
int i = 1; // 初始化条件变量
// 循环求和 1, 4, ...
while (i <= n) {
res += i;
// 更新条件变量
i++;
i *= 2;
}
return res;
}
```
=== "Zig"
@@ -724,7 +754,23 @@ status: new
=== "C"
```c title="iteration.c"
[class]{}-[func]{nestedForLoop}
/* 双层 for 循环 */
char *nestedForLoop(int n) {
// n * n 为对应点数量,"(i, j), " 对应字符串长最大为 6+10*2加上最后一个空字符 \0 的额外空间
int size = n * n * 26 + 1;
char *res = malloc(size * sizeof(char));
// 循环 i = 1, 2, ..., n-1, n
for (int i = 1; i <= n; ++i) {
// 循环 j = 1, 2, ..., n-1, n
for (int j = 1; j <= n; ++j) {
char tmp[26];
snprintf(tmp, sizeof(tmp), "(%d, %d), ", i, j);
strncat(res, tmp, size - strlen(res) - 1);
}
}
return res;
}
```
=== "Zig"
@@ -910,7 +956,16 @@ status: new
=== "C"
```c title="recursion.c"
[class]{}-[func]{recur}
/* 递归 */
int recur(int n) {
// 终止条件
if (n == 1)
return 1;
// 递:递归调用
int res = recur(n - 1);
// 归:返回结果
return n + res;
}
```
=== "Zig"
@@ -1091,7 +1146,14 @@ status: new
=== "C"
```c title="recursion.c"
[class]{}-[func]{tailRecur}
/* 尾递归 */
int tailRecur(int n, int res) {
// 终止条件
if (n == 0)
return res;
// 尾递归调用
return tailRecur(n - 1, res + n);
}
```
=== "Zig"
@@ -1278,7 +1340,16 @@ status: new
=== "C"
```c title="recursion.c"
[class]{}-[func]{fib}
/* 斐波那契数列:递归 */
int fib(int n) {
// 终止条件 f(1) = 0, f(2) = 1
if (n == 1 || n == 2)
return n - 1;
// 递归调用 f(n) = f(n-1) + f(n-2)
int res = fib(n - 1) + fib(n - 2);
// 返回结果 f(n)
return res;
}
```
=== "Zig"