函数和指针,递归调用 1.不能死递归,都要有跳出条件,每次递归逼近跳出条件 2.递归层次不能太深 3.求n个斐波拉契数

#define _CRT_SECURE_NO_WARNINGS 1 #include <stdio.h> #include <string.h> get_max(int x,int y) { int z = 0; if (x > y) { z = x; } else { z = y; } return z; } int main() { int a = 0; int b = 0; scanf("%d%d", &a, &b); int max = get_max(a, b); printf("%d", max); return 0; } 无法完成交换 void Swap(int x, int y) { int z = 0; z = x; x = y; y = z; } int main() { int a = 10; int b = 20; //写函数,交换2个整形变量的值 printf("交换前:a=%d b=%d\n", a, b); Swap(a, b); printf("交换后:a=%d b=%d\n", a, b); } int main() { int a = 0; int* pa = &a; *pa = 20; printf("%d", a); return 0; } void Swap2(int* pa, int* pb) { int z = 0; z = *pa; *pa = *pb; *pb = z; } int main() { int a= 10; int b = 20; //写函数,交换2个整形变量的值 Swap1(a, b);//传值调用 printf("交换前:a=%d b=%d\n", a, b); Swap2(&a, &b);//传址调用 printf("交换后:a=%d b=%d\n", a, b); } 函数判断100-200 素数 int is_prime(int n ) { int j = 0; for (j = 2; j < n; j++) { if (n % j == 0) { return 0; } } return 1; } int is_prime(int n) { int j = 0; for (j = 2; j <=sqrt(n); j++) { if (n % j == 0) { return 0; } } return 1; } int main() { int i = 0; int count = 0; for (i = 100; i <= 200; i++) { if (is_prime(i) == 1) { count++; printf("%d", i); } } printf("\n%d\n", count); return 0; } 判断1000-2000的闰年 判断是闰年返回1 不是闰年,返回0 int is_leap_year(int n) { if ((n % 4 == 0) && (n % 100 != 0) || (n % 400 == 0)) { return 1; } else { return 0; } } int main() { int y = 0; int count = 0; for (y = 1000; y <= 2000; y++) { if (is_leap_year(y) == 1) { count++; printf("%d", y); } } printf("\ncount=%d\n", count); return 0; } 传过来的地址名可以一样,也可以不一样 int binary_search(int a[], int k, int s) { int left = 0; int right = s - 1; while (left<=right) { int mid = (left + right) / 2; if (a[mid] > k) { right = mid - 1; } else if (a[mid] < k) { left = mid + 1; } else { return mid; } } return -1;//找不到 } int main() { int arr[] = { 1,2,3,4,5,6,7,8,9,10 }; int key = 7; int sz = sizeof(arr) / sizeof(arr[10]); //找到了就返回找的位置的下标 //找不到就返回-1 int ret = binary_search(arr, key, sz); if (-1 == ret) { printf("找不到\n"); } else { printf("找到了,下标是:%d\n", ret); } return 0; } num ++ void Add(int* p) { (*p)++; } int main() { int num = 0; Add(&num); printf("%d\n", num); Add(&num); printf("%d\n", num); Add(&num); printf("%d\n", num); } int main() { printf("%d", printf("%d", printf("%d", 43))); return 0; } int main() { int a = 10; int b = 20; //函数声明一下,告知 int Add(int, int); int c = Add(a, b); printf("%d\n", c); return 0; } int Add(int x, int y) { return x + y; } //函数递归 int main() { printf("hehe\n"); main(); return 0; } 接受一个整型值(无符号), 按照顺序打印它的每一位.例如:输入1234,打印1 2 3 4 void print(unsigned int n) { if (n > 9) { print(n / 10); } printf("%d", n % 10); } int main() { unsigned int num = 0; scanf("%u", &num); //递归,函数自己调用自己 print(num);//print函数可以打印参数部分数字的每一位 return 0; } 写递归代码的时候 1.不能死递归,都要有跳出条件,每次递归逼近跳出条件 2.递归层次不能太深 int main() { char arr[] = "bit"; printf("%d\n", strlen(arr)); return 0; } int my_strlen(char* str) { int count = 0; while (*str != '\0') { count++; str++; } return count; } int main() { char arr[] = "bit"; //['b'] ['i'] ['t'] ['\0'] printf("%d\n",my_strlen(arr)); return 0; } int my_strlen(char* str) { if (*str != '\0') return 1 + my_strlen(str+1); else return 0; } int main() { char arr[] = "bit"; //['b'] ['i'] ['t'] ['\0'] printf("%d\n", my_strlen(arr)); return 0; } n阶乘 int main() { int n = 0; scanf("%d", &n); int i = 0; int ret = 1; for (i = 1; i <= n; i++) { ret = ret * i; } printf("%d\n", ret); return 0; } 递归方法写n阶乘 int Fac(int n) { if (n <= 1) { return 1; } else { return n * Fac(n - 1); } } int main() { int n = 0; scanf("%d", &n); int ret = Fac(n); printf("%d\n", ret); return 0; } 求n个斐波拉契数 1,1,2,3,5,8,13,21,34,55 int count = 0; //递归可以求解,但是效率太低 int Fib(int n) { //统计第三个斐波拉契的计算次数 if (n == 3) { count++; } if (n <= 2) { return 1; } else { return Fib(n - 1) + Fib(n - 2); } } int main() { int n = 0; scanf("%d", &n); int ret = Fib(n); printf("%d\n", ret); printf("count=%d\n", count); return 0; } 算法优化 int Fib(int n) { int a = 1; int b = 1; int c = 1; while (n > 2) { c = a + b; a = b; b = c; n--; } return c; } int main() { int n = 0; scanf("%d", &n); int ret = Fib(n); printf("%d\n", ret); return 0; }