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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#462770#7087. Counting PolygonsFFTiltedWA 280ms198552kbC++206.2kb2024-07-04 03:39:322024-07-04 03:39:33

Judging History

你现在查看的是最新测评结果

  • [2024-07-04 03:39:33]
  • 评测
  • 测评结果:WA
  • 用时:280ms
  • 内存:198552kb
  • [2024-07-04 03:39:32]
  • 提交

answer

#include "bits/stdc++.h"

#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(a) (a).begin(), (a).end()
#define ar array

using namespace std;

using vi = vector<int>;
using ll = long long;
using pi = pair<int, int>;

const ll INF = 7e18;

template<typename T>
bool ckmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}


using ll = long long;

const int MOD = 1e9 + 7;

struct mint {
    int value;

    mint(ll x = 0) {
        if (x >= MOD || x < 0) {
            x %= MOD;
            if (x < 0) x += MOD;
        }
        value = x;
    }

    mint &operator+=(const mint &b) {
        value += b.value;
        if (value >= MOD) value -= MOD;
        return *this;
    }

    mint &operator-=(const mint &b) {
        value -= b.value;
        if (value < 0) value += MOD;
        return *this;
    }

    mint &operator*=(const mint &b) {
        value = (1ll * value * b.value) % MOD;
        return *this;
    }

    mint inv();
};

mint operator*(mint a, const mint &b) {
    return a *= b;
}

mint operator+(mint a, const mint &b) {
    return a += b;
}

mint operator-(mint a, const mint &b) {
    return a -= b;
}

mint power(mint a, ll b) {
    mint r = 1;
    for (; b; b >>= 1, a *= a) if (b & 1) r *= a;
    return r;
}

mint mint::inv() {
    return power(*this, MOD - 2);
}


void solve() {
    constexpr int N = 1e7, M = 2e7;
    vector<int> minp(N + 1);
    vector<mint> fact(M + 1), rfct(M + 1);

    fact[0] = 1;
    for (int i = 1; i <= M; ++i) fact[i] = i * fact[i - 1];
    rfct[M] = fact[M].inv();
    for (int i = M; i >= 1; --i) rfct[i - 1] = rfct[i] * i;

    for (int i = 2; i <= N; ++i) {
        if (minp[i]) continue;
        minp[i] = i;
        if ((long long) i * i > N) continue;
        for (int j = i * i; j <= N; j += i) {
            if (!minp[j]) minp[j] = i;
        }
    }

    auto c = [&](int n, int k) {
        if (0 <= k && k <= n)
            return fact[n] * rfct[k] * rfct[n - k];
        else return mint(0);
    };

    auto T = [&](int n, int k) {
        return c(n - 1, k - 1);
    };

    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;

        mint ans = 0;
        int big = (n - 1) / 2;
        {
            {
                // identity
                ans += T(n, m);
                ans -= m * T(n - big, m);
            }

            vector<pair<int, int>> prs;
            for (int x = m; x != 1;) {
                int p = minp[x];
                int cp = 0;
                while (x % p == 0) {
                    cp++;
                    x /= p;
                }
                prs.emplace_back(p, cp);
            }

            vector<int> divs;
            function<void(int, int)> rec = [&](int i, int d) {
                if (i == divs.size()) {
                    divs.push_back(d);
                    return;
                }
                rec(i + 1, d);
                for (int j = 1; j <= prs[i].second; ++j) {
                    d *= prs[i].first;
                    rec(i + 1, d);
                }
            };
            rec(0, 1);
            sort(all(divs));
            vector<int> cnt(divs.size());
            for (int i = (int) divs.size() - 1; i >= 0; --i) {
                int g = divs[i];
                cnt[i] = m / g;
                for(int j = i + 1; j < divs.size(); ++j) {
                    if (divs[j] % g == 0) {
                        cnt[i] -= cnt[j];
                    }
                }
            }
            cnt.back()--;


            rep(i, divs.size()) {
                int g = divs[i];
                if (g == m) continue;
//                cerr << g << ' ' << cnt[i] << '\n';
                int parts = m / g;
                assert(parts >= 2);
                if (n % parts != 0) {
                    continue;
                }
                int m1 = g;
                int n1 = n / parts;
                ans += T(n1, m1) * mint(cnt[i]);
            }
//            cerr << "END\n";
        }
        {
            // reverse
            // f(i) = (n - 1 - i + s) % n
            auto solve = [&](int s) {
                if (m % 2 == 1) {
                    // unique solution
                    int pairs = (m - 1) / 2;
                    int ones = 1;
                    mint value = T((n + 1) / 2, pairs + ones);
                    value -= T((n - big + 1) / 2, pairs + ones);
                    return value;
                } else if ((s - 1) % 2 != 0) {
                    // no solution
                    int pairs = m / 2;
                    if (n % 2 != 0) return mint(0);
                    return T(n / 2, pairs);
                } else {
                    // two solutions
                    int pairs = (m - 2) / 2;
                    int ones = 2;
                    mint value = 0;
                    rep(mask, 4) {
                        int q = __builtin_popcount(mask);
                        if ((q + n) % 2 == 0) {
                            value += T((n + q) / 2, pairs + ones);
                            if (big % 2 == 0) {
                                value -= T((n + q - big) / 2, pairs + ones) * 2;
                            } else {
                                rep(t, 2) {
                                    if (mask & (1 << t)) {
                                        value -= T((n + q - (big + 1)) / 2, pairs + ones);
                                    } else {
                                        value -= T((n + q - (big - 1)) / 2, pairs + ones);
                                    }
                                }
                            }
                        }
                    }
                    return value;
                }
            };
            if (m % 2 == 1) {
                ans += solve(0) * m;
            } else {
                ans += (solve(1) + solve(0)) * (m / 2);
            }
        }
        ans *= mint(m * 2).inv();
        cout << ans.value << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 247ms
memory: 198552kb

input:

4
3 3
4 3
5 3
7 4

output:

1
0
1
3

result:

ok 4 tokens

Test #2:

score: -100
Wrong Answer
time: 280ms
memory: 198512kb

input:

10000
8708700 6531525
8062080 4031040
7068600 2356200
3659040 332640
7309575 2088450
5503680 1572480
7162848 3581424
9831360 7724640
6306300 5045040
5783400 2891700
4677120 2338560
5110560 786240
9525600 2381400
6955200 4173120
8229375 3291750
7068600 5405400
9639000 3213000
5969040 2984520
5274360 ...

output:

307362178
976948825
121335228
239821667
588888788
65316729
381830419
938936888
202691627
350184868
342540860
707612387
42866986
355634738
823761415
768040953
308869940
701018487
449728784
297941976
325896441
474609111
171404054
404643399
884006845
282832572
379567594
845275948
388249819
317309081
96...

result:

wrong answer 1st words differ - expected: '70575804', found: '307362178'