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计算n的阶乘
int main() { int i=0; int ret=1; scanf("%d",&i); for(i=1;i<=n;i++) {
ret=ret*i; } printf("%d\n",ret); return 0; }
计算1！+2！+3！+.....+10!
int main() {计算1！+2！+3！+.....+10! int i = 0; int j = 0; int sum = 0; int ret =
1; for (i = 1; i <= 3; i++) { ret = 1; for (j = 1; j <= i; j++) { ret = ret *
j; }sum += ret; } printf("%d\n", sum); return 0; }//算法效率低 //效率高的方法 int main() {
int n=0; int sum = 0; int ret = 1;//不可赋初值为0,0乘以任何数都为0 for(n=1;n<=10;n++) {
ret*=n;//阶乘 sum+=ret;//求和 } printf("%d\n",sum); reurn 0; }
在一个有序数组中查找具体的某个数字n
折半查找/二分查找(前提条件是数组有序）
left mid right
arr[mid]中间元素
int main() { int arr[] = { 1,2,3,4,5,6,7,8,9,10 }; int
k=10;//查找k,k的值可以是数组中的任意数 int sz = sizeof(arr) / sizeof(arr[0]);//40/4 int left
= 0; int right = sz - 1;//-1是因为数组下标从0开始 while (left <= right)//不能漏掉= { int mid
= left + (right-left) / 2;//中间元素必须放到循环里面 //int mid = (left + right) / 2; if
(arr[mid] < k) { left = mid + 1; } else if (arr[mid] > k) { right = mid - 1; }
else { printf("找到了下标是：%d\n", mid); break; } } if (left > right) {
printf("找不到了\n"); } return 0; }
编写代码，演示多个字符从两端向中间汇聚
#include<string.h> #include<windows.h> #include<stdio.h> int main() {
//下标从0开始且减去\0所以得减去2 //sizeof(arr1)/sizeof(arr[0])-2;//4 //char arr[] = "abc";
//[a b c \0] char arr1[] = "welcome to bit!!!!!"; char arr2[] =
"###################"; int left = 0; //int right =
sizeof(arr1)/sizeof(arr[0])-2; int right strlen(arr1)-1; while (left <= right)
{ arr2[left] = arr1[left]; arr2[right] = arr1[right]; printf("%s\n",arr2);
Sleep(1000);//单位是毫秒 system("cls"); left++; right--; } printf("%s\n",arr2);
return 0; }
模拟用户登录情景
密码最多输入三次,，正确则提示输入成功，输入三次错误就退出程序
两个字符的比较不能使用==，应该使用strcmp
#include<string.h> #include<stdio.h> int main() { int i = 0; char password[20]
= {0}; //假设正确密码是“abcdef” for(i = 0; i < 3; i++) { printf("请输入密码：");
scanf("%s",passsword); if(strcmp(password,"abcdef")==0) { printf("密码正确\n");
break; } else { printf("密码错误，重新输入\n"); } } if (i == 3) {
printf("三次密码均错误，退出程序\n"); } return 0; }