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Mr.Lee每隔1/x s攻击一次,cpu每隔1/y s攻击一次 因为时间与答案无关,最后只看boss受到了多少次攻击 所以可以在每个人的频率上同时乘以xy 即Mr.Lee每隔y s攻击一次,cpu每隔x s攻击一次 这样看虽然时间延长但是结果不变 就可以二分查找出打败boss用时,最后再根据时间判断谁给予的最后一击 二分出用时t,则t%x==0表示cpu给予最后一击 t%y==0表示Mr.Lee给予最后一击 #include<stdio.h> int main(){ long long n,…
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这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Game of Stones 题目来源 第10届IEEE极限编程大赛 https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/game-of-stones-1-1 Alice and Bob play a game. The game is turn based: Alice moves first, then Bob, a…