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ctfshow-大赛原题(680-685)
web680 post传入code=phpinfo();有回显 查看disable_functions，发现这些被禁用 assert,system,passthru,exec,pcntl_exec,shell_exec,popen,proc_open,pcntl_alarm,pcntl_fork,pcntl_waitpid,pcntl_wait,pcntl_wifexited,pcntl_wifstoped,pcntl_wifsignaled,pcntl_wexitstatus,pcn685 vue3 mixin，extends
认识Mixin Mixin的基本使用 Mixin的合并规则 全局混入Mixin extends import { createApp } from 'vue'; import App from './01_mixin和extends/App.vue'; // import App from './02_compositionAPI基础/App.vue'; const app = createApp(App); // 全局Codeforces Round #685 Div. 2 C 题解
从第一步操作可以得到程序的结果与字母的顺序无关, 所以不妨转化为数组z[26][2]来储存a,b字符串字母的数量。 而第二步操作中 如果直接按照题设的步骤来做， 即对a字符串进行“k个相同字母+1” 那么写出的代码会十分繁琐，且不容易打出正确的代码。 所以我们不妨逆向操作 对bCodeforces Round #685 (Div. 2) A. Subtract or Divide 构造
Ridbit starts with an integer n. In one move, he can perform one of the following operations: divide n by one of its proper divisors, or subtract 1 from n if n is greater than 1. A proper divisor is a divisor of a number, excluding itself. For example, 1Codeforces Round #685 (Div. 2)
Codeforces Round #685 (Div. 2) A - Subtract or Divide 枚举 int main() { IOS; for (cin >> _; _; --_) { cin >> n; int ans = 0; if (n == 1) cout << 0 << '\n'; else if (n == 2) cout << 1 << '\n';