結果

問題 No.1043 直列大学
ユーザー firiexp
提出日時 2020-05-01 22:02:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 2,533 bytes
コンパイル時間 962 ms
コンパイル使用メモリ 100,040 KB
最終ジャッジ日時 2025-01-10 04:50:03
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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ファイルパターン 結果
sample AC * 3
other AC * 13 WA * 15
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 <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
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, m;
    cin >> n >> m;

    vector<int> dp1(100001), dp2(100001);
    dp1[0] = 1; dp2[0] = 1;
    for (int i = 0; i < n; ++i) {
        int x; cin >> x;
        for (int j = 100000-x; j >= 0; --j) {
            dp1[j+x] += dp1[j];
        }
    }
    for (int i = 0; i < m; ++i) {
        int x; cin >> x;
        for (int j = 100000-x; j >= 0; --j) {
            dp2[j+x] += dp2[j];
        }
    }
    dp1[0] = 0; dp2[0] = 0;
    int L, R;
    cin >> L >> R;
    cout << [&](int L, int R){
        int l = 0, r = 0; mint val = 0, ans = 0;
        for (int i = 0; i <= 100000; ++i) {
            while(r <= 100000 && r*R < i) val -= dp2[r++];
            while(l <= 100000 && l*L <= i) val += dp2[l++];
            ans += val*dp1[i];
        }
        return ans;
    }(L, R).val << "\n";
    return 0;
}
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