65 lines
1.6 KiB
C++
65 lines
1.6 KiB
C++
#include <iostream>
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#include <bitset>
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#include <type_traits>
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using ull = unsigned long long;
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using std::cout,std::iostream,std::bitset;
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template<class T>
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class ReadNumber{
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char c;
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T w,n;
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public:
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ReadNumber& operator>>(T &num)noexcept{
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c=(char)0,w=1,n=0;
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while(!isdigit(c)){
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if constexpr(!std::is_same_v<ull, T>){
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if(c=='-')w=-1;
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}
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c = getchar();
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}
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while(isdigit(c)){
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n = n*10 + (c-'0');
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c = getchar();
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}
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num = w*n;
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return *this;
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}
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};
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ull n, m, c, k, num, input, p, ans {1}, _;
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bitset<64+5> bt;
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ReadNumber<ull> readull;
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#define cin readull
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template<class T>
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constexpr T safe_left_shift(const T t,const size_t n)noexcept{
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constexpr auto sizeof_t = sizeof(t)*8;
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return (n>=sizeof_t?0:t<<n);
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}
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int main(){
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cin>>n>>m>>c>>k;
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if(n==0 && m==0 && k==64){ //特判
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cout<<"18446744073709551616\n";
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return 0;
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}
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for(ull i=0;i<n;i++){
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cin>>input;
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num |= input; //所有编号占用的'1'的位置，把他并在一起
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}
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for(ull i=0;i<m;i++){
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cin>>p>>_;
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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
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k--;
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bt[p]=true;
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}
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}
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// 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
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ans = safe_left_shift(1ull, k); // 避开未定义行为
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cout<<ans-n<<'\n';//注意这里要去重，基本排列组合原理
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} |