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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
#405483	#7613. Inverse Problem	ucup-team3215	TL	1ms	3572kb	C++20	2.1kb	2024-05-06 00:41:28	2024-05-06 00:41:30

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你现在查看的是最新测评结果

[2024-05-06 00:41:30]
评测

测评结果：TL
用时：1ms
内存：3572kb

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[2024-05-06 00:41:28]
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answer

#include <algorithm>
#include <cstdint>
#include <iostream>
#include <numeric>
#include <unordered_map>
#include <vector>

using namespace std;

constexpr int mod = 1e9 + 7, cut = 6;

auto& mul(auto&& a, auto b) { return a = a * (uint64_t) b % mod; }
int inv(int a) { for (int r = 1, b = mod - 2; ; b /= 2, mul(a, a)) if (!b) return r; else if (b % 2) mul(r, a); }

using namespace std;

vector<unordered_map<int, vector<int>>> high(1);
vector<int> inv_(1);

void gen_high(int n0, int n, int c, int p, auto&& cur) {
  high[n0 - n].emplace(p, cur);
  for (; c > cut; --c) if (c <= n) {
    int np = p;
    for (int i = 0; i < c; ++i) mul(np, inv_[n0 - i]);
    cur.push_back(c);
    gen_high(n0, n - c, c, np, cur);
    cur.pop_back();
  }
}

vector<int> qs;
vector<vector<int>> ans;

void solve(int n0, int n, int c, int p, auto&& cur) {
  for (int i = 0; i < qs.size(); ++i) if (ans[i].empty() && high[n].count(mul(+p, qs[i]))) {
    ans[i] = cur;
    auto& t = high[n][mul(+p, qs[i])];
    ans[i].insert(ans[i].end(), t.begin(), t.end());
  }
  for (; c > 0; --c) if (c <= n) {
    int np = p;
    for (int i = 0; i < c; ++i) mul(np, n0 - i);
    cur.push_back(c);
    solve(n0, n - c, c, np, cur);
    cur.pop_back();
  }
}

int main() {
  int tc; cin >> tc;
  qs.resize(tc);
  ans.resize(tc);
  for (int i = 0; i < tc; ++i) {
    cin >> qs[i];
    if (qs[i] == 1) ans[i] = {-1};
    else if (qs[i] == 2) ans[i] = {-1};
    qs[i] = inv(qs[i]);
  }
  for (int n = 1, done = 0; !done++; ++n) {
    inv_.push_back(inv(n));
    high.push_back({});
    for (auto& x: high) x.clear(), x.max_load_factor(.2);
    gen_high(n, n, n, 1, vector<int>{});
    solve(n, n, cut, mul(n + 1, n + 2), vector<int>{});
    for (int i = 0; i < tc; ++i) done &= !ans[i].empty();
  }
  for (int i = 0; i < tc; ++i) {
    if (qs[i] == 1) cout << "1\n";
    else {
      if (ans[i][0] == -1) ans[i] = {};
      cout << accumulate(ans[i].begin(), ans[i].end(), 0) + 2 << '\n' << "1 2\n";
      for (int j = 0, k = 2; j < ans[i].size(); ++j)
      while (ans[i][j]--) cout << j + 1 << ' ' << ++k << '\n';
    }
  }
}

源文件, 语言: C++20

詳細信息

Test #1:

score: 100

Accepted

time: 1ms

memory: 3572kb

input:

output:

result:

ok OK (4 test cases)

Test #2:

score: -100

Time Limit Exceeded

input:

output: