algorithm_2024/P7076/P7076.cpp

57 lines
1.3 KiB
C++
Raw Normal View History

2024-10-03 15:29:49 +00:00
#include <iostream>
2024-10-03 16:10:57 +00:00
#include <bitset>
#include <type_traits>
2024-10-03 15:29:49 +00:00
2024-10-03 16:10:57 +00:00
using ull = unsigned long long;
2024-10-03 16:24:14 +00:00
using std::cout,std::iostream,std::bitset;
2024-10-03 16:10:57 +00:00
template<class T>
class ReadNumber{
char c;
T w,n;
public:
ReadNumber& operator>>(T &num)noexcept{
c=(char)0,w=1,n=0;
while(!isdigit(c)){
if constexpr(!std::is_same_v<ull, T>){
if(c=='-')w=-1;
}
c = getchar();
}
while(isdigit(c)){
n = n*10 + (c-'0');
c = getchar();
}
num = w*n;
return *this;
}
};
2024-10-03 16:24:14 +00:00
ull n,m,c,k,num,input,p,q,ans {1};
bitset<64+5> bt;
2024-10-03 16:10:57 +00:00
ReadNumber<ull> readull;
#define cin readull
2024-10-02 06:29:36 +00:00
int main(){
2024-10-03 15:29:49 +00:00
cin>>n>>m>>c>>k;
2024-10-03 16:10:57 +00:00
2024-10-03 16:51:00 +00:00
if(n==0 && m==0 && k==64){ //特判
2024-10-03 16:10:57 +00:00
cout<<"18446744073709551616\n";
return 0;
}
for(ull i=0;i<n;i++){
cin>>input;
2024-10-03 16:51:00 +00:00
num |= input; //所有编号占用的'1'的位置,把他并在一起
2024-10-03 16:10:57 +00:00
}
for(ull i=0;i<m;i++){
cin>>p>>q;
2024-10-03 16:51:00 +00:00
if(!((num>>p)&1) && !bt[p]){ //已有的n个编号里面没有出现过的规则并且还未标记,bt用来去重
2024-10-03 16:10:57 +00:00
k--;
bt[p]=true;
}
}
2024-10-03 16:51:00 +00:00
for(int i=0;i<k;i++)ans*=2; //此时ans=所有(未排除n个已用规则),可以用的二进制位的组合
cout<<ans-n<<'\n';//注意这里要去重
2024-10-02 06:29:36 +00:00
}