結果

問題 No.1043 直列大学
ユーザー hirono999hirono999
提出日時 2020-05-01 22:09:48
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 168 ms / 2,000 ms
コード長 6,225 bytes
コンパイル時間 2,233 ms
コンパイル使用メモリ 211,672 KB
実行使用メモリ 164,016 KB
最終ジャッジ日時 2024-06-02 02:53:07
合計ジャッジ時間 5,855 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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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 12 ms
10,768 KB
testcase_01 AC 7 ms
9,212 KB
testcase_02 AC 24 ms
15,424 KB
testcase_03 AC 8 ms
9,208 KB
testcase_04 AC 8 ms
9,212 KB
testcase_05 AC 18 ms
9,216 KB
testcase_06 AC 18 ms
9,212 KB
testcase_07 AC 11 ms
12,316 KB
testcase_08 AC 21 ms
12,316 KB
testcase_09 AC 94 ms
85,904 KB
testcase_10 AC 110 ms
101,572 KB
testcase_11 AC 145 ms
150,916 KB
testcase_12 AC 168 ms
164,016 KB
testcase_13 AC 133 ms
137,624 KB
testcase_14 AC 139 ms
145,364 KB
testcase_15 AC 152 ms
158,720 KB
testcase_16 AC 136 ms
142,244 KB
testcase_17 AC 107 ms
105,444 KB
testcase_18 AC 108 ms
105,504 KB
testcase_19 AC 129 ms
134,532 KB
testcase_20 AC 104 ms
103,076 KB
testcase_21 AC 118 ms
124,260 KB
testcase_22 AC 122 ms
129,716 KB
testcase_23 AC 151 ms
162,596 KB
testcase_24 AC 110 ms
115,652 KB
testcase_25 AC 129 ms
135,928 KB
testcase_26 AC 139 ms
145,300 KB
testcase_27 AC 79 ms
73,344 KB
testcase_28 AC 135 ms
134,384 KB
testcase_29 AC 51 ms
46,764 KB
testcase_30 AC 104 ms
102,576 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
const long long INF = 1LL << 60;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= (n); ++i)
#define rrep(i, n) for (ll i = (n - 1); i >= 0; --i)
#define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c)))
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define pb push_back
#define to_s to_string
#define len(v) (ll)v.size()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define print(x) cout << (x) << '\n'
#define drop(x) cout << (x) << '\n', exit(0)
#define debug(x) cout << #x << ": " << (x) << '\n'
using namespace std;
using ll = long long;
typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vector<ll>> vec2;
typedef vector<vector<vector<ll>>> vec3;
template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); }
inline ll devc(ll x, ll y) { return (x + y - 1) / y; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }

struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;

template< typename T1, typename T2 >
ostream &operator << (ostream &os, const pair< T1, T2 > &p) {
    os << p.first << " " << p.second;
    return os;
}

template< typename T1, typename T2 >
istream &operator >> (istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}

template< typename T1, typename T2, typename T3 >
ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) {
    os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t);
    return os;
}

template< typename T1, typename T2, typename T3 >
istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) {
    is >> get<0>(t) >> get<1>(t) >> get<2>(t);
    return is;
}

template< typename T >
ostream &operator << (ostream &os, const vector< T > &v){
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}

template< typename T >
istream &operator >>  (istream &is, vector< T > &v){
    for(T &in : v) is >> in;
    return is;
}

/*---------------------------------    Tools    ------------------------------------------*/
template< typename T >
vector<T> cumsum(const vector<T> &X){
    vector<T> res(X.size() + 1, 0);
    for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i];
    return res;
}

template< typename S, typename T, typename F>
pair<T, T> bisearch(S left, T right, F f) {
    while(abs(right - left) > 1){
        T mid = (right + left) / 2;
        if(f(mid)) right = mid;
        else left = mid;
    }
    return {left, right};
}

template< typename S, typename T, typename F>
double trisearch(S left, T right, F f, int maxLoop = 90){
    double low = left, high = right;
    while(maxLoop--){
        double mid_left = high / 3 + low * 2 / 3;
        double mid_right = high * 2 / 3 + low / 3;
        if(f(mid_left) >= f(mid_right)) low = mid_left;
        else high = mid_right;
    }
    return (low + high) * 0.5;
}

//Def of Monoid
//Suppose that S is a set and ● is some binary opeartion S x S -> S
//then S with ● is a monoid if it satisfies the following two:
// Associativity(結合則)
//  For all a,b and c in S, the equation (a ● b) ● c = a ● (b ● c) holds
// Identitiy element(単位元の存在)
//  There exisits an element e in S such that for every element a in S,
//  the equations e ● a = a ● e = a holds

//Eample of Monoid
//+, *, and, or, xor, min, max

//Build O(N)
//Query O(log N)
//- query(a,b) : applay operation to the range [a, b)
//- update(k,x) : change k-th element to x
//- operator[k] : return k-th element

template< typename Monoid >
class SegmentTree{
    private:
        using F = function<Monoid(Monoid, Monoid)>;
        
        long long sz;
        vector<Monoid> seg;
        const F f;
        const Monoid e;

    public:
        SegmentTree(long long n, const F f, const Monoid &e) : f(f), e(e){
            sz = 1;
            while(sz < n) sz <<= 1;
            seg.assign(2 * sz, e);
        }

        void set(long long k, const Monoid &x){
            seg[k + sz] = x;
        }

        void build() {
            for (long long k = sz - 1; k > 0; --k){
                seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
            }
        }

        void update(long long k, const Monoid &x) {
            k += sz;
            seg[k] = x;
            while(k >>= 1) seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }

        Monoid query(long long a, long long b){
            Monoid L = e, R = e;
            for (a += sz, b += sz; a < b; a >>= 1, b >>= 1){
                if(a & 1) L = f(L, seg[a++]);
                if(b & 1) R = f(seg[--b],R);
            }
            return f(L, R);
        }

        Monoid operator[](const int &k) const {
            return seg[k + sz];
        }
};

/*------------------------------- Main Code Here -----------------------------------------*/

int main()
{
    ll N, M;
    cin >> N >> M;
    vec V(N), R(M);
    cin >> V >> R;
    ll A, B;
    cin >> A >> B;

    vec2 f(N + 1, vec(1e5 + 1, 0));
    vec2 g(M + 1, vec(1e5 + 1, 0));
    f[0][0] = g[0][0] = 1;

    rep1(i, N) rep(j, 1e5 + 1){
        f[i][j] = f[i - 1][j];
        if(j >= V[i - 1]) f[i][j] += f[i - 1][j - V[i - 1]], f[i][j] %= MOD;
    }
    rep1(i, M) rep(j, 1e5 + 1){
        g[i][j] = g[i - 1][j];
        if(j >= R[i - 1]) g[i][j] += g[i - 1][j - R[i - 1]], g[i][j] %= MOD;
    }

    SegmentTree<ll> segtree(1e5 + 10, [&](ll x, ll y){ return (x + y) % MOD; }, 0);

    rep(i, 1e5 + 1) segtree.set(i, f[N][i]);
    segtree.build();

    ll ans = 0;

    for (ll r = 1; r <= 1e5; ++r){
        ans += g[M][r] * segtree.query(A * r, min(B * r + 1, ll(1e5)));
        ans %= MOD;
    }
    print(ans);

    return 0;
}
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