计算n的阶乘
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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!
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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 |
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//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
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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
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#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
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#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; } |
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原文链接:https://blog.csdn.net/Ll_R_lL/article/details/122524132