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However, I do notice that to make the dark situation happen, it doesn't require the topleft matrix to be realized. It could even be some matrix as below at minimum, Human,    AIHuman      no        noAI         yes/no     no(or yes doesn't even matte…
E. The Last Fight Between Human and AI 题目连接: http://codeforces.com/contest/676/problem/E Description 100 years have passed since the last victory of the man versus computer in Go. Technologies made a huge step forward and robots conquered the Earth!…
链接 Codeforces 676E The Last Fight Between Human and AI 题意 给一个多项式,有些系数不确定.人和机器轮流填系数,系数可以是任何数,问是否能使得最后的多项式整除 x-k 思路 整除x-k就等价 x=k时,多项式为0. 系数可以是任何数的话就能得到任何结果,只要人是否是填最后一个就行. 另外分 k=0 和 k<>0两种情况考虑. 代码 #include <iostream> #include <cstdio> #incl…
人类和电脑在一个多项式上进行博弈,多项式的最高次项已知,一开始系数都不确定.电脑先开始操作,每次操作可以确定某次项的系数,这个系数可以是任意实数.给出一个博弈中间状态,最后如果这个多项式被x-K整除就算人类赢,问人类是否有可能赢.n<=1e5,K和所有给出的系数的绝对值在1e4内,不确定的系数用?表示. 这个读入有点坑爹... 要使,做一下除法,余数大概长这个样子 问题即有没有可能使L为0. (一)K=0时,决胜关键就在a0,如果a0确定了并且不为0那就完了,如果a0没确定,看轮到谁,轮到人类赢…
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Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
贪心 A Nicholas and Permutation #include <bits/stdc++.h> typedef long long ll; const int N = 1e5 + 5; int a[105]; int pos[105]; int main() { int n; scanf ("%d", &n); for (int i=1; i<=n; ++i) { scanf ("%d", a+i); pos[a[i]] =…