1 条题解

  • 2
    @ 2024-3-1 20:20:09
    #include<bits/stdc++.h>
    using namespace std;
    using i128 = __int128_t;
    const int OFFSET = 1;
    const int N = 5e7 + OFFSET, P = 6002268 + OFFSET;
    const int MOD = 1e9 + 7;
    bool vis[N];
    int f[N];
    int prime[P], cnt_prime;
    constexpr i128 pow6(int k) {
    k = (1ll * k * k) % MOD;
    return (i128)k * k * k % MOD;
    }
    void init_prime(const int &n = N - 1) {
    f[1] = 1;
    for (int i = 2; i <= n; ++i) {
    if (!vis[i]) f[prime[++cnt_prime] = i] = pow6(i) - 1;
    for (int j = 1; j <= cnt_prime && i * prime[j] <= n; ++j) {
    vis[i * prime[j]] = 1;
    f[i * prime[j]] = f[i] * pow6(prime[j]) % MOD;
    if (f[i * prime[j]] >= MOD) f[i * prime[j]] -= MOD;
    if (i % prime[j] == 0) break;
    if ((f[i * prime[j]] -= f[i]) < 0) f[i * prime[j]] += MOD;
    }
    }
    for (int i = 2; i <= n; ++i)
    if ((f[i] += f[i - 1]) >= MOD) f[i] -= MOD;
    }
    auto __STATIC__ = []() {
    init_prime();
    return 0.0;
    }();
    constexpr int F(const int x,
    const int d,
    const int r,
    const int e,
    const int a,
    const int m) {
    return (i128)(d / x) * (r / x) * (e / x) * (a / x) % MOD * (m / x) % MOD;
    }
    int _m;
    i128 ans;
    int d, r, e, a, m;
    int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int _t;
    cin >> _t;
    while (_t--) {
    cin >> d >> r >> e >> a >> m;
    _m = min({d, r, e, a, m});
    ans = 0;
    for (int _l = 1, _r; _l <= _m; _l = _r + 1) {
    _r = min(
    {d / (d / _l), r / (r / _l), e / (e / _l), a / (a / _l), m / (m / _l)});
    ans += 1ll * (f[_r] - f[_l - 1] + MOD) * F(_l, d, r, e, a, m) % MOD;
    }
    cout << (int)(ans % MOD) << '\n';
    }
    return 0;
    }
    

    信息

    ID
    172
    时间
    5000ms
    内存
    512MiB
    难度
    7
    标签
    (无)
    递交数
    151
    已通过
    31
    上传者