結果
問題 | No.213 素数サイコロと合成数サイコロ (3-Easy) |
ユーザー | heno239 |
提出日時 | 2020-08-15 14:18:42 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 13 ms / 3,000 ms |
コード長 | 4,723 bytes |
コンパイル時間 | 1,624 ms |
コンパイル使用メモリ | 125,524 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-10-10 17:53:04 |
合計ジャッジ時間 | 2,079 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 8 ms
5,248 KB |
testcase_01 | AC | 13 ms
5,248 KB |
ソースコード
#include<iostream> #include<string> #include<cstdio> #include<vector> #include<cmath> #include<algorithm> #include<functional> #include<iomanip> #include<queue> #include<ciso646> #include<random> #include<map> #include<set> #include<bitset> #include<stack> #include<unordered_map> #include<utility> #include<cassert> #include<complex> #include<numeric> using namespace std; //#define int long long typedef long long ll; typedef unsigned long long ul; typedef unsigned int ui; constexpr ll mod =1000000007; const ll INF = mod * mod; typedef pair<int, int>P; #define stop char nyaa;cin>>nyaa; #define rep(i,n) for(int i=0;i<n;i++) #define per(i,n) for(int i=n-1;i>=0;i--) #define Rep(i,sta,n) for(int i=sta;i<n;i++) #define rep1(i,n) for(int i=1;i<=n;i++) #define per1(i,n) for(int i=n;i>=1;i--) #define Rep1(i,sta,n) for(int i=sta;i<=n;i++) #define all(v) (v).begin(),(v).end() typedef pair<ll, ll> LP; typedef double ld; typedef pair<ld, ld> LDP; const ld eps = 1e-12; const ld pi = acos(-1.0); ll mod_pow(ll x, ll n, ll m = mod) { ll res = 1; while (n) { if (n & 1)res = res * x % m; x = x * x % m; n >>= 1; } return res; } struct modint { ll n; modint() :n(0) { ; } modint(ll m) :n(m) { if (n >= mod)n %= mod; else if (n < 0)n = (n % mod + mod) % mod; } operator int() { return n; } }; bool operator==(modint a, modint b) { return a.n == b.n; } modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; } modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; } modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; } modint operator+(modint a, modint b) { return a += b; } modint operator-(modint a, modint b) { return a -= b; } modint operator*(modint a, modint b) { return a *= b; } modint operator^(modint a, ll n) { if (n == 0)return modint(1); modint res = (a * a) ^ (n / 2); if (n % 2)res = res * a; return res; } ll inv(ll a, ll p) { return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p); } modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); } ll gcd(ll a, ll b) { if (a < b)swap(a, b); while (b) { ll r = a % b; a = b; b = r; } return a; } struct kitamasa { private: vector<ll> ori; int k; public: kitamasa(vector<ll> d) { k = d.size(); rep(i, k) { ori.push_back(d[i]); } } void plus1(vector<ll>& s) { vector<ll> ss; ss.push_back(s[k - 1] * ori[0]); rep1(i, k - 1) { ss.push_back((s[i - 1] + s[k - 1] * ori[i]) % mod); } rep(i, k)s[i] = ss[i]; } void multi2(vector<ll>& s) { vector<ll> memo = s; vector<vector<ll>> v; rep(i, k) { plus1(s); v.push_back(s); } rep(i, k) { ll csum = 0; rep(j, k) { csum += memo[j] * v[j][i] % mod; csum %= mod; } s[i] = csum; } } vector<ll> calc(ll n) { n += k; vector<ll> s; s.push_back(1); rep(i, k - 1)s.push_back(0); ll cop = n; int cnt = 0; while (cop) { cnt++; cop >>= 1; } per(i, cnt - 1) { ll z = (ll)1 << i; multi2(s); if (n & z) { plus1(s); } } return s; } }; vector<int> a = { 2,3,5,7,11,13 }; vector<int> b = { 4,6,8,9,10,12 }; modint dp1[3901][6]; modint sum1[3901]; modint dp2[3601][6]; modint sum2[3601]; modint dp[7501]; void solve() { ll n; cin >> n; int p, c; cin >> p >> c; dp1[0][0] = 1; rep(i, p) { per(j, 13 * i + 1) { per(k, 6) { for (int l = k; l < 6; l++) { dp1[j + a[l]][l] += dp1[j][k]; } dp1[j][k] = 0; } } } dp2[0][0] = 1; rep(i, c) { per(j, 12*i + 1) { per(k, 6) { for (int l = k; l < 6; l++) { dp2[j + b[l]][l] += dp2[j][k]; } dp2[j][k] = 0; } } } rep(i, 13 * p + 1) { rep(j, 6)sum1[i] += dp1[i][j]; } rep(i, 12 * c + 1)rep(j, 6)sum2[i] += dp2[i][j]; rep(i, 13 * p + 1)rep(j, 12 * c + 1)dp[i + j] += sum1[i] * sum2[j]; int sup = 13 * p + 12 * c; vector<ll> ori(sup); rep(i, sup) { ori[i] = dp[sup-i]; } kitamasa kt(ori); ll le = max((ll)0, (ll)n - sup); //cout << "?? " << le << "\n"; vector<ll> s = kt.calc(le); int sz = min((ll)sup, n); vector<ll> val(sz); rep(i, sz) { val[i] = s.back(); kt.plus1(s); //rep(j, s.size())cout << s[j] << " "; cout << "\n"; } //cout << le << " " << s.back() << "\n"; modint ans = 0; for (int i = sz; i < sz + sup; i++) { rep(j, sup+1) { int to = i - j; if (to >= 0 && to < sz) { ans += dp[j] * (modint)val[to]; } } } /*rep(i, sup + 1)cout <<i<<" "<< sum1[i]<<" "<<sum2[i] << " "<<dp[i]<<"\n"; rep(i, sz) { cout << val[i] << "\n"; }*/ cout << ans << "\n"; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(15); //init_f(); //init(); //expr(); //int t; cin >> t; rep(i, t) solve(); return 0; }