CF778B(round 402 div.2 E) Bitwise Formula
题意:
Bob recently read about bitwise operations used in computers: AND, OR and XOR. He have studied their properties and invented a new game.
Initially, Bob chooses integer m, bit depth of the game, which means that all numbers in the game will consist of mbits. Then he asks Peter to choose some m-bit number. After that, Bob computes the values of n variables. Each variable is assigned either a constant m-bit number or result of bitwise operation. Operands of the operation may be either variables defined before, or the number, chosen by Peter. After that, Peter's score equals to the sum of all variable values.
Bob wants to know, what number Peter needs to choose to get the minimum possible score, and what number he needs to choose to get the maximum possible score. In both cases, if there are several ways to get the same score, find the minimum number, which he can choose.
The first line contains two integers n and m, the number of variables and bit depth, respectively (1 ≤ n ≤ 5000; 1 ≤ m ≤ 1000).
The following n lines contain descriptions of the variables. Each line describes exactly one variable. Description has the following format: name of a new variable, space, sign ":=", space, followed by one of:
- Binary number of exactly m bits.
- The first operand, space, bitwise operation ("AND", "OR" or "XOR"), space, the second operand. Each operand is either the name of variable defined before or symbol '?', indicating the number chosen by Peter.
Variable names are strings consisting of lowercase Latin letters with length at most 10. All variable names are different.
In the first line output the minimum number that should be chosen by Peter, to make the sum of all variable values minimum possible, in the second line output the minimum number that should be chosen by Peter, to make the sum of all variable values maximum possible. Both numbers should be printed as m-bit binary numbers.
3 3
a := 101
b := 011
c := ? XOR b
011
100
5 1
a := 1
bb := 0
cx := ? OR a
d := ? XOR ?
e := d AND bb
0
0
思路:
1.由于是位操作,所以各个位之间的运算结果相互独立。因此一位一位割裂来计算,选取每一位取“0”还是取“1”更好一点。对于每一位来讲,n个数的这一位上的“1”的个数的和越多,则所有数的和越大。另外,如果取“0”和取“1”能获得相同的结果,则取“0”即可。
2.学习使用bitset用法。
实现:
#include <iostream>
#include <cstdio>
#include <bitset>
#include <map>
using namespace std; int n, m, res[][];
bitset<> bs[][];
map<string, int> mp;
int main()
{
mp["?"] = ;
bs[][].set();
cin >> n >> m;
for (int i = ; i < n; i++)
{
string x, tmp, y, op, z;
cin >> x >> tmp >> y;
mp[x] = i + ;
if (y[] == '' || y[] == '')
{
for (int j = ; j < m; j++)
{
bs[i + ][].set(j, y[j] - '');
bs[i + ][].set(j, y[j] - '');
}
}
else
{
cin >> op >> z;
for (int j = ; j <= ; j++)
{
if (op == "AND")
{
bs[i + ][j] = bs[mp[y]][j] & bs[mp[z]][j];
}
else if (op == "OR")
{
bs[i + ][j] = bs[mp[y]][j] | bs[mp[z]][j];
}
else
{
bs[i + ][j] = bs[mp[y]][j] ^ bs[mp[z]][j];
}
}
}
for (int j = ; j < m; j++)
{
for (int k = ; k <= ; k++)
res[j][k] += bs[i + ][k][j];
}
}
for (int i = ; i < m; i++)
{
cout << (res[i][] > res[i][] ? : );
}
puts("");
for (int i = ; i < m; i++)
{
cout << (res[i][] >= res[i][] ? : );
}
return ;
}
CF778B(round 402 div.2 E) Bitwise Formula的更多相关文章
- 【推导】【贪心】Codeforces Round #402 (Div. 2) E. Bitwise Formula
按位考虑,每个变量最终的赋值要么是必为0,要么必为1,要么和所选定的数相同,记为2,要么和所选定的数相反,记为3,一共就这四种情况. 可以预处理出来一个真值表,然后从前往后推导出每个变量的赋值. 然后 ...
- Codeforces Round #402 (Div. 2)
Codeforces Round #402 (Div. 2) A. 日常沙比提 #include<iostream> #include<cstdio> #include< ...
- Codeforces Round #402 (Div. 2) A+B+C+D
Codeforces Round #402 (Div. 2) A. Pupils Redistribution 模拟大法好.两个数列分别含有n个数x(1<=x<=5) .现在要求交换一些数 ...
- Codeforces Round #402 (Div. 2) A,B,C,D,E
A. Pupils Redistribution time limit per test 1 second memory limit per test 256 megabytes input stan ...
- CodeForces Round #402 (Div.2) A-E
2017.2.26 CF D2 402 这次状态还算能忍吧……一路不紧不慢切了前ABC(不紧不慢已经是在作死了),卡在D,然后跑去看E和F——卧槽怎么还有F,早知道前面做快点了…… F看了看,不会,弃 ...
- Codeforces Round #402 (Div. 1)
A题卡壳了,往离线倒着加那方面想了会儿,后来才发现方向错了,二十多分钟才过掉,过了B后做D,想法好像有点问题,最后只过两题,掉分了,差一点回紫. AC:AB Rank:173 Rating:2227- ...
- Codeforces Round#402(Div.1)掉分记+题解
哎,今天第一次打div1 感觉头脑很不清醒... 看到第一题就蒙了,想了好久,怎么乱dp,倒过来插之类的...突然发现不就是一道sb二分吗.....sb二分看了二十分钟........ 然后第二题看了 ...
- Codeforces Round #402 (Div. 2) A B C sort D二分 (水)
A. Pupils Redistribution time limit per test 1 second memory limit per test 256 megabytes input stan ...
- 【DFS】Codeforces Round #402 (Div. 2) B. Weird Rounding
暴搜 #include<cstdio> #include<algorithm> using namespace std; int n,K,Div=1,a[21],m,ans=1 ...
随机推荐
- 数据结构之 图论---bfs(邻接表)
数据结构实验之图论二:基于邻接表的广度优先搜索遍历 Time Limit: 1000MS Memory limit: 65536K 题目描述 给定一个无向连通图,顶点编号从0到n-1,用广度优先搜索( ...
- get all sites under IIS
https://stackoverflow.com/questions/2555668/how-to-programmatically-get-sites-list-and-virtual-dirs- ...
- “There's no Qt version assigned to this project for platform ” - visual studio plugin for Qt
1.find menu "Qt VS Tools", select Qt Options 2.add a new Qt version 3. right click the tar ...
- windows server2012之部署HTTPS安全站点
现在的互联网越来越重视网络安全方面的内容,像我们日常生活中浏览的网上银行网站等涉及安全的你都会发现有https 的标志出现,在URL前加https://前缀表明是用SSL加密的. 你的电脑与服务器之间 ...
- 从OutStreamWriter 和Filewriter谈Java编码
首先看JAVA API的描述: ABOUT OutputStreamWriter: "An OutputStreamWriter is a bridge from character str ...
- 【HDU 4722】 Good Numbers
[题目链接] 点击打开链接 [算法] f[i][j]表示第i位,数位和对10取模余j的数的个数 状态转移,计算答案都比较简单,笔者不再赘述 [代码] #include<bits/stdc++.h ...
- ES6 模板编译
顾名思义,就是用反引号编写一个模板字符串, 用echo将模板转为javascrip表达式字符串, 用正则将基础字符串转为想要字符串 将代码封装在函数中返回: 注: 用到es6属性${} var tem ...
- YARN(MapReduce 2)运行MapReduce的过程-源码分析
这是我的分析,当然查阅书籍和网络.如有什么不对的,请各位批评指正.以下的类有的并不完全,只列出重要的方法. 如要转载,请注上作者以及出处. 一.源码阅读环境 需要安装jdk1.7.0版本及其以上版本, ...
- hdoj1027【STL系列。。。?】
这个太夸张了...感觉是有别的方法,但是觉得再说吧...以后碰到全排列应该也是用STL嗨的吧...嗯,,,就是这样的....?再说,再说.. 还有杭电支持c艹11,很棒 #include <bi ...
- bzoj 3513: [MUTC2013]idiots【生成函数+FFT】
想了好长时间最后发现真是石乐志 第一反应就是两边之和大于第三边,但是这个东西必须要满足三次-- 任意的两边之和合通过生成函数套路+FFT求出来(记得去掉重复选取的),然后这任意两边之和大于任意第三边可 ...