#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

题目链接:http://www.hihocoder.com/problemset/problem/1636 题目描述 在中国古代神话中,盘古是时间第一个人并且开天辟地,它从混沌中醒来并把混沌分为天地. 刚开始地上是没有山的,只有满地的石头. 这里有 \(N\) 堆石头,标号为从 \(1\) 到 \(N\) .盘古想要把它们合成一堆建造一座大山.如果某些堆石头的数量总和是 \(S\) ,盘古需要 \(S\) 秒才能把它们合成一堆,这新的一堆石头的数量就是 \(S\) . 不幸的是,每一次盘古只能把…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the beginning, t…

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 859    Accepted Submission(s): 325 Problem Description Every school has some legends, Northeastern University is the same. Enter…

In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the beginning, there was no mountain on the earth, only stones all over…

题意:有一块n*n的田,田上有一些点可以放置稻草人,再给出一些稻草人,每个稻草人有其覆盖的距离ri,距离为曼哈顿距离,求要覆盖到所有的格子最少需要放置几个稻草人 由于稻草人数量很少,所以状态压缩枚举,之后慢慢判断即可,注意放稻草人的格子是不需要覆盖的 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include&…

题目链接 题意 :小女孩注册了两个比赛的帐号,初始分值都为0,每做一次比赛如果排名在前两百名,rating涨50,否则降100,告诉你她每次比赛在前两百名的概率p,如果她每次做题都用两个账号中分数低的那个去做,问她最终有一个账号达到1000分需要做的比赛的次数的期望值. 思路 :可以直接用公式推出来用DP做,也可以列出210个方程组用高斯消元去做. (1)DP1:离散化.因为50,100,1000都是50的倍数,所以就看作1,2,20.这样做起来比较方便. 定义dp[i]为从 i 分数到达i+1…

https://www.luogu.org/problemnew/show/P3147 此题与上一题完全一样,唯一不一样的就是数据范围; 上一题是248,而这一题是262144; 普通的区间dp表示状态表示法根本存不下, 这时我们就要想另外的状态表示法; #include <bits/stdc++.h> #define read read() #define up(i,l,r) for(int i = (l);i <=(r); i++) using namespace std; int…