2016-2017 ACM-ICPC Northeastern European Regional Contest (NEERC 16) Gym - 101190H bitset，逻辑表达式模拟

题目链接：http://codeforces.com/gym/101190/attachments

题意：给了一堆逻辑表达式，判断是否对于所有的整数(-2^15 - 2^15-1)范围类的数，都满足逻辑表达式或者都不满足这些逻辑表达式，还有是部分满足，部分满足需要输出这些区间。没说清楚，题目挺短的，可以自己读下。

解法： bitset暴力，注意把下标修正到正数，设个offset就好。

#include<bits/stdc++.h>
using namespace std;
const int offset = 32768;
int main()
{
freopen("hard.in","r",stdin);
freopen("hard.out","w",stdout);
string str;
bitset<65536> ans;
while(getline(cin, str))
{
stringstream ss(str);
string s;
bitset<65536> tmp;
tmp.flip();
while(1)
{
ss >> s;
ss >> s;
bitset<65536> make;
if(s == ">=")
{
ss >> s;
int x;
sscanf(s.c_str(), "%d", &x);
for(int i = x + offset; i < 65536; i++) make.set(i);
}
else
{
ss >> s;
int x;
sscanf(s.c_str(), "%d", &x);
for(int i = x + offset; i >= 0; i--) make.set(i);
}
tmp &= make;
if(!(ss >> s)) break;
if(s == "||") break;
}
ans |= tmp;
}
if(ans.count() == 65536)
{
puts("true");
return 0;
}
else if(ans.count() == 0)
{
puts("false");
return 0;
}
vector<pair<int, int> > Ans;
for(int i = 0; i < 65536; i++)
{
if(ans[i])
{
int pos = i;
while(pos + 1 < 65536 && ans.test(pos + 1)) pos++;
Ans.push_back(make_pair(i, pos));
i = pos;
}
}
for(int i = 0; i < Ans.size(); i++)
{
if(Ans[i].first == 0 || Ans[i].second == 65535)
{
if(Ans[i].first == 0)
printf("x <= %d", Ans[i].second - offset);
else
printf("x >= %d", Ans[i].first - offset);
}
else
printf("x >= %d && x <= %d", Ans[i].first - offset, Ans[i].second - offset);
if(i != Ans.size() - 1) printf(" ||");
printf("\n");
}
}