結果
問題 | No.573 a^2[i] = a[i] |
ユーザー |
|
提出日時 | 2020-06-26 16:17:19 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 335 ms / 2,000 ms |
コード長 | 2,543 bytes |
コンパイル時間 | 1,190 ms |
コンパイル使用メモリ | 125,760 KB |
最終ジャッジ日時 | 2025-01-11 10:36:26 |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 47 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:115:20: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 115 | ll n; scanf("%lld", &n); | ~~~~~^~~~~~~~~~~~
ソースコード
#include <cstdio>#include <iostream>#include <string>#include <sstream>#include <stack>#include <algorithm>#include <cmath>#include <queue>#include <map>#include <set>#include <cstdlib>#include <bitset>#include <tuple>#include <assert.h>#include <deque>#include <bitset>#include <iomanip>#include <limits>#include <chrono>#include <random>#include <array>#include <unordered_map>#include <functional>#include <complex>#include <numeric>template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }constexpr long long MAX = 5100000;constexpr long long INF = 1LL << 60;constexpr int inf = 1000000007;constexpr long long mod = 1000000007LL;//constexpr long long mod = 998244353LL;const long double PI = acos((long double)(-1));using namespace std;typedef unsigned long long ull;typedef long long ll;typedef long double ld;struct mint {long long x;mint(long long x = 0) :x((x% mod + mod) % mod) {}mint& operator+=(const mint a) {if ((x += a.x) >= mod) x -= mod;return *this;}mint& operator-=(const mint a) {if ((x += mod - a.x) >= mod) x -= mod;return *this;}mint& operator*=(const mint a) {(x *= a.x) %= mod;return *this;}mint operator+(const mint a) const {mint res(*this);return res += a;}mint operator-(const mint a) const {mint res(*this);return res -= a;}mint operator*(const mint a) const {mint res(*this);return res *= a;}mint pow(ll t) const {if (!t) return 1;mint a = pow(t >> 1);a *= a;if (t & 1) a *= *this;return a;}// for prime modmint inv() const {return pow(mod - 2);}mint& operator/=(const mint a) {return (*this) *= a.inv();}mint operator/(const mint a) const {mint res(*this);return res /= a;}};long long fac[MAX], finv[MAX], inv[MAX];void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++) {fac[i] = fac[i - 1] * i % mod;inv[i] = mod - inv[mod % i] * (mod / i) % mod;finv[i] = finv[i - 1] * inv[i] % mod;}}mint COM(int n, int k) {if (n < k) return 0;if (n < 0 || k < 0) return 0;return mint(fac[n] * (finv[k] * finv[n - k] % mod) % mod);}int main(){/*cin.tie(nullptr);ios::sync_with_stdio(false);*/COMinit();ll n; scanf("%lld", &n);mint res = 0;for (int i = 0; i <= n; i++) {res += COM(n, i) * mint(n - i).pow(i);}cout << res.x << "\n";return 0;}