bzoj前100题计划

xz布置的巨大的坑。。

有空填题解。。。


1002 轮状病毒

用python手动matrixtree打表。

#include<bits/stdc++.h>
#define il inline
#define vd void
typedef long long ll;
il int gi(){
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
char biao[101][45]={"mmp","1","5","16","45","121","320","841","2205","5776","15125","39601","103680","271441","710645","1860496","4870845","12752041","33385280","87403801","228826125","599074576","1568397605","4106118241","10749957120","28143753121","73681302245","192900153616","505019158605","1322157322201","3461452808000","9062201101801","23725150497405","62113250390416","162614600673845","425730551631121","1114577054219520","2918000611027441","7639424778862805","20000273725560976","52361396397820125","137083915467899401","358890350005878080","939587134549734841","2459871053643326445","6440026026380244496","16860207025497407045","44140595050111976641","115561578124838522880","302544139324403592001","792070839848372253125","2073668380220713167376","5428934300813767249005","14213134522220588579641","37210469265847998489920","97418273275323406890121","255044350560122222180445","667714778405043259651216","1748099984655007556773205","4576585175559979410668401","11981655542024930675232000","31368381450514812615027601","82123488809519507169850805","215002084978043708894524816","562882766124611619513723645","1473646213395791149646646121","3858055874062761829426214720","10100521408792494338631998041","26443508352314721186469779405","69230003648151669220777340176","181246502592140286475862241125","474509504128269190206809383201","1242282009792667284144565908480","3252336525249732662226888342241","8514727565956530702536099118245","22291846172619859445381409012496","58360810951903047633608127919245","152790586683089283455442974745241","400010949097364802732720796316480","1047242260609005124742719414204201","2741715832729650571495437446296125","7177905237579946589743592924684176","18791999880010189197735341327756405","49198094402450621003462431058585041","128802283327341673812651951847998720","337208755579574400434493424485411121","882823983411381527490828321608234645","2311263194654570182037991540339292816","6050965600552329018623146299409643805","15841633607002416873831447357889638601","41473935220454921602871195774259272000","108580172054362347934782139964888177401","284266580942632122201475224120405260205","744219570773534018669643532396327603216","1948392131377969933807455373068577549445","5100956823360375782752722586809405045121","13354478338703157414450712387359637585920","34962478192749096460599414575269507712641","91532956239544131967347531338448885552005","239636390525883299441443179440077148943376","627376215338105766356982006981782561278125",};
int main(){
#ifndef ONLINE_JUDGE
freopen("1002.in","r",stdin);
freopen("1002.out","w",stdout);
#endif
printf("%s",biao[gi()]);
return 0;
}
from decimal import Decimal,getcontext
getcontext().prec=80
eps=Decimal("0."+"0"*50+"1")
for n in range(3,101):
A=[[Decimal('0') for i in range(0,n)]for i in range(0,n)]
for i in range(0,n):A[i][i]=Decimal('3')
for i in range(0,n):
for j in range(0,n):
if i!=j:A[i][j]=Decimal(str(-int(abs(i-j)==1 or abs(i-j)==n-1)))
for i in range(0,n):
for j in range(i+1,n):
if(abs(A[j][i])>abs(A[i][i])):
A[i],A[j]=A[j],A[i]
for j in range(i+1,n):
if abs(A[j][i]-Decimal('0'))<eps:continue
S=A[j][i]/A[i][i]
for k in range(i,n):
A[j][k]-=A[i][k]*S
ans=Decimal('1')
for i in range(0,n):ans*=A[i][i]
print("\""+str(int(ans))+"\"",end=',')

Decimal高精度真tm好


1004 cards

神仙Burnside。玄学求出所有洗牌法都没有不动点(除了单位元)

所以答案是不动点个数/(m+1),不动点个数是把Srgb组合一下


1005 明明的烦恼

怎么全是高精好烦啊

就是prufer序列上面d[i]!=-1的点会出现d[i]-1次,设前cnt个点d[i]!=-1,\(sum=\sum_{i=1}^{cnt}d[i]-1\)

所以这部分是\(\frac{sum!}{\Pi_{i=1}^{cnt} (d[i]-1)!}\)

剩下的点一共有\(n-cnt\)个,出现\(n-2-sum\)次,所以这部分是\((n-cnt)^{n-2-sum}\)

然后还要乘一个\(C_{n-2}^{sum}\)

所以最后答案就是\(C_{n-2}^{sum} \times \frac{sum!}{\Pi_{i=1}^{cnt} (d[i]-1)!}\times (n-cnt)^{n-2-sum}\)

还能约个分$ \frac{(n-2)!(n-cnt){n-2-sum}}{(n-2-sum)!\Pi_{i=1}{cnt} (d[i]-1)!}$

这个诡异的东西,高精求就行了。。。


1037 [ZJOI2008]生日聚会Party

f[i][j][k][l] 表示选i个♂的j个♀的,右边每一段最多男比女多k女比男多l的方案数


1038 瞭望塔

每条直线的上半平面求个交。答案一定在原来折线的交点或上凸壳的顶点上。

#include<bits/stdc++.h>
#define il inline
#define vd void
typedef long long ll;
il int gi(){
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
double x[301],y[301];
struct line{double k,b;bool del;}l[301],s[301],b[301];
int n,top;
il bool operator <(const line&a,const line&b){
if(a.del)return 0;if(b.del)return 1;
if(fabs(a.k-b.k)>1e-9)return a.k<b.k;
return a.b>b.b;
}
il double jx(const line&a,const line&b){return(a.b-b.b)/(b.k-a.k);}
il double jy(const line&a,const line&b){return(a.k*b.b-a.b*b.k)/(a.k-b.k);}
il double solve(double X,line*s,int N){
for(int i=1;i<N;++i)if(X-1e-9<=jx(s[i],s[i+1]))return s[i].k*X+s[i].b;
return s[N].k*X+s[N].b;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("1038.in","r",stdin);
freopen("1038.out","w",stdout);
#endif
n=gi();
for(int i=1;i<=n;++i)scanf("%lf",&x[i]);
for(int i=1;i<=n;++i)scanf("%lf",&y[i]);
int m=n-1;
for(int i=1;i<=m;++i){
l[i].k=b[i].k=(y[i]-y[i+1])/(x[i]-x[i+1]);
l[i].b=b[i].b=y[i]-l[i].k*x[i];
l[i].del=0;
}
std::sort(l+1,l+m+1);
for(int i=1;i<=m;++i)if(fabs(l[i].k-l[i-1].k)<1e-7)l[i].del=1;
std::sort(l+1,l+m+1);
l[m+1].del=1;m=0;while(!l[m+1].del)++m;
top=0;
s[++top]=l[1];
for(int i=2;i<=m;++i){
while(top>1&&jx(s[top],l[i])<=jx(s[top-1],l[i])-1e-9)--top;
s[++top]=l[i];
}
double ans=1e18;
ans=std::min(ans,solve(x[1],s,top)-solve(x[1],b,n-1));
ans=std::min(ans,solve(x[n],s,top)-solve(x[n],b,n-1));
for(int i=1;i<n;++i){
double p=jx(b[i],b[i+1]);
ans=std::min(ans,solve(p,s,top)-solve(p,b,n-1));
}
for(int i=1;i<top;++i){
double p=jx(s[i],s[i+1]);
if(p+1e-9>=x[1]&&p<=x[n]+1e-9)ans=std::min(ans,solve(p,s,top)-solve(p,b,n-1));
}
if(ans<0)ans=0;
printf("%.3lf\n",ans);
return 0;
}

bzoj 前100题计划的更多相关文章

  1. 【总结】 BZOJ前100题总结

    前言 最近发现自己trl,所以要多做题目但是Tham布置的题目一道都不会,只能来写BZOJ HA(蛤)OI 1041 复数可以分解成两个点,所以直接把\(R^2\)质因数分解一下就可以了,注意计算每一 ...

  2. LeetCode前100题(EASY难度)

    1 Two Sum Given an array of integers, return indices of the two numbers such that they add up to a s ...

  3. 纪念,BZOJ AC 100题!

    虽然说有将近50+是usaco,然后还有很多水题T_T 看来我还是刷水题.... 看来我还是那么弱. T_T 但是好歹也要留个纪念..

  4. BZOJ第一页刷题计划

    BZOJ第一页刷题计划 已完成:67 / 90 [BZOJ1000]A+B Problem:A+B: [BZOJ1001][BeiJing2006]狼抓兔子:最小割: [BZOJ1002][FJOI2 ...

  5. bzoj 3208 花神的秒题计划I

    bzoj 3208 花神的秒题计划I Description 背景[backboard]: Memphis等一群蒟蒻出题中,花神凑过来秒题-- 描述[discribe]: 花花山峰峦起伏,峰顶常年被雪 ...

  6. csp退役前的做题计划1(真)

    csp退役前的做题计划1(真) 因为我太菜了,所以在第一次月考就会退役,还是记录一下每天做了什么题目吧. 任务计划 [ ] Z算法(Z Algorithm) 9.28 [x] ARC061C たくさん ...

  7. NOIP2018前的一些计划&记录(日更)

    先空着,等停课了再开始写. 诸位好,我是yyb.现在显然已经不再是高一的小蒟蒻了,已经升级为了高二的菜鸡了 现在已经不能再每天划划水切切题了,毕竟......已经高二了,所有的机会从高一的两倍全部除了 ...

  8. bzoj Usaco补完计划(优先级 Gold>Silver>资格赛)

    听说KPM初二暑假就补完了啊%%% 先刷Gold再刷Silver(因为目测没那么多时间刷Silver,方便以后TJ2333(雾 按AC数降序刷 ---------------------------- ...

  9. NOIP前的一些计划

    一些想法 距离NOIP2018只剩下一个星期的时间了,通过这几天在长郡的考试,渐渐感觉还有好多东西自己还不够熟练,也有些东西到现在还不会,现将NOIP前的一些计划列在这里,希望能在考前把他们全部完成吧 ...

随机推荐

  1. 使用CAReplicatorLayer [2]

    使用CAReplicatorLayer [2] 工具类 // // Math.h // MathEquation // // Created by YouXianMing on 15/11/20. / ...

  2. zabbix的日常监控-API

    Zabbix API提供两项主要功能: 远程管理Zabbix配置 远程检索配置和历史数据 官方文档:https://www.zabbix.com/documentation/3.4/zh/manual ...

  3. python第三方库——xlrd和xlwt操作Excel文件学习

    python第三方库——xlrd和xlwt操作Excel文件学习 1安装: C:\Users\Lenovo>pip install xlwtCollecting xlwt  Downloadin ...

  4. 面向对象课程 - 寒假第三次作业 - C++计算器项目初始部分

    C++计算器项目初始部分 零.项目源文件地址 传送门:calculator 一.项目信息相关: 项目:Calculator 版本:1.0 日期:2016.2.16 实现: 基本的操作界面 对四则运算表 ...

  5. chrome开发者工具那点事

    Elements:查找网页源代码HTML中的任一元素,手动修改任一元素的属性和样式且能实时在浏览器里面得到反馈. Console:记录开发者开发过程中的日志信息,且可以作为与JS进行交互的命令行She ...

  6. python伪装网页访问

    # -*- coding:utf8 -*-#import urllib.request#url =' http://www.douban.com/'#webPage=urllib.request.ur ...

  7. Lambda表达式和For循环使用需要注意的一个地方

    一个需要注意的地方看下面的代码: using System; using System.Collections.Generic; using System.Linq; namespace MyCsSt ...

  8. F2eTest程序快捷方式安装到桌面

      按照http://f2etest.***.com/install页面上介绍的:安装快捷方式到Windows桌面 (Win7)部分的介绍进行安装. 其中第三步的https连接需要在f2etest的c ...

  9. docker 导入导出镜像

    docker容器导入导出有两种方法: 一种是使用save和load命令 使用例子如下: docker save ubuntu:load>/root/ubuntu.tar docker load& ...

  10. kong安装

    1 哪些企业在用KONG: Mashapoe Marketplace Buzzlogix Gengo IBM Intel OpenDNS 饿了么 Kong可与两种不同的组件协同工作: Nginx:Ko ...