結果
| 問題 |
No.641 Team Contest Estimation
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-03-26 11:16:09 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
CE
(最新)
AC
(最初)
|
| 実行時間 | - |
| コード長 | 2,590 bytes |
| コンパイル時間 | 688 ms |
| コンパイル使用メモリ | 86,084 KB |
| 最終ジャッジ日時 | 2024-11-14 22:13:01 |
| 合計ジャッジ時間 | 1,540 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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コンパイルエラー時のメッセージ・ソースコードは、提出者また管理者しか表示できないようにしております。(リジャッジ後のコンパイルエラーは公開されます)
ただし、clay言語の場合は開発者のデバッグのため、公開されます。
ただし、clay言語の場合は開発者のデバッグのため、公開されます。
コンパイルメッセージ
main.cpp:18:39: error: '::numeric_limits' has not been declared
18 | template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;
| ^~~~~~~~~~~~~~
main.cpp:18:55: error: expected primary-expression before '>' token
18 | template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;
| ^
main.cpp:18:61: error: no matching function for call to 'max()'
18 | template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;
| ~~~~~^~
In file included from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/string:50,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/bits/locale_classes.h:40,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/bits/ios_base.h:41,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ios:42,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/ostream:38,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/iostream:39,
from main.cpp:1:
/home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/bits/stl_algobase.h:254:5: note: candidate: 'template<class _Tp> constexpr const _Tp& std::max(const _Tp&, const _Tp&)'
254 | max(const _Tp& __a, const _Tp& __b)
| ^~~
/home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/bits/stl_algobase.h:254:5: note: template argument deduction/substitution failed:
main.cpp:18:61: note: candidate expects 2 arguments, 0 provided
18 | template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;
| ~~~~~^~
/home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/bits/stl_algobase.h:300:5: note: candidate: 'templat
ソースコード
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>
static const int MOD = 1000000009;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;
template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;
template<u32 M = 1000000007>
struct modint{
u32 val;
modint(): val(0){}
template<typename T>
modint(T t){t %= (T)M; if(t < 0) t += (T)M; val = t;}
modint pow(ll k) const {
modint res(1), x(val);
while(k){
if(k&1) res *= x;
x *= x;
k >>= 1;
}
return res;
}
template<typename T>
modint& operator=(T t){t %= (T)M; if(t < 0) t += (T)M; val = t; return *this;}
modint inv() const {return pow(M-2);}
modint& operator+=(modint a){val += a.val; if(val >= M) val -= M; return *this;}
modint& operator-=(modint a){if(val < a.val) val += M-a.val; else val -= a.val; return *this;}
modint& operator*=(modint a){val = (u64)val*a.val%M; return *this;}
modint& operator/=(modint a){return (*this) *= a.inv();}
modint operator+(modint a) const {return modint(val) +=a;}
modint operator-(modint a) const {return modint(val) -=a;}
modint operator*(modint a) const {return modint(val) *=a;}
modint operator/(modint a) const {return modint(val) /=a;}
modint operator-(){return modint(M-val);}
bool operator==(const modint a) const {return val == a.val;}
bool operator!=(const modint a) const {return val != a.val;}
bool operator<(const modint a) const {return val < a.val;}
};
using mint = modint<MOD>;
int main() {
int n, k;
cin >> n >> k;
vector<int> cnt1(k);
for (int i = 0; i < n; ++i) {
ll x;
scanf("%lld", &x);
for (int j = 0; j < k; ++j) {
if(x & (1LL << j)){
cnt1[j]++;
}
}
}
mint Ex, Exx, tmp = 1;
mint A = mint(2).pow(k)-mint(1);
Ex = mint(n)*A*mint(500000005);
Exx = A*A;
for (int i = 0; i < k; ++i) {
Exx -= tmp*tmp;
tmp += tmp;
}
Exx *= mint(n)*mint(n)*mint(750000007);
tmp = 1;
mint tmp2;
for (int i = 0; i < k; ++i) {
tmp2 += tmp*tmp*(mint(cnt1[i]).pow(2)+mint(n-cnt1[i]).pow(2));
tmp += tmp;
}
Exx += tmp2*mint(500000005);
mint ans1 = Ex*(mint(2).pow(k));
mint ans2 = (Exx-Ex*Ex)*(mint(4).pow(k));
cout << ans1.val << "\n";
cout << ans2.val << "\n";
return 0;
}