結果
問題 | No.590 Replacement |
ユーザー | 夕叢霧香(ゆうむらきりか) |
提出日時 | 2017-11-04 00:27:38 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,981 bytes |
コンパイル時間 | 1,288 ms |
コンパイル使用メモリ | 105,484 KB |
実行使用メモリ | 18,176 KB |
最終ジャッジ日時 | 2024-11-23 15:31:36 |
合計ジャッジ時間 | 5,532 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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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 | 2 ms
6,816 KB |
testcase_01 | AC | 1 ms
6,816 KB |
testcase_02 | AC | 2 ms
6,820 KB |
testcase_03 | AC | 2 ms
6,816 KB |
testcase_04 | AC | 2 ms
6,820 KB |
testcase_05 | AC | 2 ms
6,816 KB |
testcase_06 | AC | 2 ms
6,816 KB |
testcase_07 | AC | 7 ms
6,816 KB |
testcase_08 | AC | 4 ms
6,820 KB |
testcase_09 | AC | 8 ms
6,816 KB |
testcase_10 | AC | 4 ms
6,816 KB |
testcase_11 | AC | 2 ms
6,820 KB |
testcase_12 | AC | 8 ms
6,820 KB |
testcase_13 | AC | 37 ms
6,816 KB |
testcase_14 | AC | 55 ms
6,816 KB |
testcase_15 | AC | 5 ms
6,816 KB |
testcase_16 | AC | 13 ms
6,816 KB |
testcase_17 | AC | 23 ms
6,816 KB |
testcase_18 | AC | 86 ms
7,508 KB |
testcase_19 | AC | 86 ms
6,912 KB |
testcase_20 | AC | 21 ms
6,816 KB |
testcase_21 | AC | 103 ms
7,688 KB |
testcase_22 | AC | 103 ms
9,180 KB |
testcase_23 | AC | 110 ms
8,296 KB |
testcase_24 | AC | 109 ms
8,012 KB |
testcase_25 | AC | 112 ms
7,680 KB |
testcase_26 | AC | 111 ms
7,808 KB |
testcase_27 | AC | 112 ms
8,280 KB |
testcase_28 | AC | 110 ms
7,424 KB |
testcase_29 | AC | 112 ms
7,424 KB |
testcase_30 | AC | 114 ms
7,168 KB |
testcase_31 | AC | 116 ms
8,136 KB |
testcase_32 | AC | 115 ms
8,368 KB |
testcase_33 | AC | 2 ms
6,820 KB |
testcase_34 | AC | 2 ms
6,820 KB |
testcase_35 | AC | 2 ms
6,816 KB |
testcase_36 | AC | 88 ms
9,504 KB |
testcase_37 | AC | 88 ms
9,500 KB |
testcase_38 | AC | 83 ms
11,848 KB |
testcase_39 | AC | 83 ms
11,848 KB |
testcase_40 | AC | 90 ms
11,536 KB |
testcase_41 | AC | 91 ms
11,668 KB |
testcase_42 | AC | 82 ms
10,808 KB |
testcase_43 | AC | 80 ms
9,636 KB |
testcase_44 | AC | 124 ms
18,176 KB |
testcase_45 | WA | - |
testcase_46 | WA | - |
ソースコード
#include <algorithm> #include <cassert> #include <iostream> #include <map> #include <vector> using namespace std; typedef long long lint; typedef pair<int, int> pii; const lint MOD = 1000000007; const bool DEBUG = false; int gcd(int x, int y) { while (y != 0) { int remainder = x % y; x = y; y = remainder; } return x; } vector<int> inverse(const vector<int> &permutation) { int n = permutation.size(); vector<int> result(n); for (int i = 0; i < n; ++i) { result[permutation[i]] = i; } return result; } void cycles(const vector<int> &permutation, vector<int> &cycle_id, vector<int> &period, vector<int> &order) { int n = permutation.size(); cycle_id = vector<int>(n, -1); order = vector<int>(n); vector<int> periods; int count = 0; for (int i = 0; i < n; ++i) { if (cycle_id[i] >= 0) { continue; } cycle_id[i] = count; order[i] = 0; int p = 1; int current = permutation[i]; while (current != i) { cycle_id[current] = count; order[current] = p; current = permutation[current]; p++; } periods.push_back(p); count++; } period = periods; } // http://www.geeksforgeeks.org/chinese-remainder-theorem-set-2-implementation/ // Returns modulo inverse of a with respect to m using extended // Euclid Algorithm. Refer below post for details: // http://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/ lint inv(lint a, lint m) { lint m0 = m, t, q; lint x0 = 0, x1 = 1; if (m == 1) return 0; // Apply extended Euclid Algorithm while (a > 1) { // q is quotient q = a / m; t = m; // m is remainder now, process same as // euclid's algo m = a % m, a = t; t = x0; x0 = x1 - q * x0; x1 = t; } // Make x1 positive if (x1 < 0) x1 += m0; return x1; } // k is size of num[] and rem[]. Returns the smallest // number x such that: // x % num[0] = rem[0], // x % num[1] = rem[1], // .................. // x % num[k-2] = rem[k-1] // Assumption: Numbers in num[] are pairwise coprime // (gcd for every pair is 1) lint findMinX(lint num[], lint rem[], int k) { // Compute product of all numbers lint prod = 1; for (int i = 0; i < k; i++) prod *= num[i]; // Initialize result lint result = 0; // Apply above formula for (int i = 0; i < k; i++) { lint pp = prod / num[i]; result += rem[i] * inv(pp, num[i]) * pp; } return result % prod; } lint crt(int x, int y, int ma, int mb, int g) { lint num[2] = {ma / g, mb / g}; lint rem[2] = {x / g, y / g}; lint sol = findMinX(num, rem, 2); lint value = sol; lint prod = (lint) ma * (mb / g); value = (value * g + (x % g)); return value % prod; } lint doit(const vector<pii> &data, int ape, int bpe, int p_gcd, int diff) { if (data.empty()) { return 0; } if (DEBUG) { cerr << "doit:" << ape << " " << bpe << endl; for (auto v: data) { cerr << v.first << " " << v.second << endl; } } lint prod = (lint) ape * (lint) bpe / p_gcd; vector<lint> tmp; for (int i = 0; i < (int) data.size(); ++i) { pii d = data[i]; int x = d.first; int y = d.second; lint value = crt(x, (y + diff) % bpe, ape, bpe, p_gcd); tmp.push_back(value); } sort(tmp.begin(), tmp.end()); tmp.push_back(tmp[0] + prod); if (DEBUG) { cerr << "tmp:"; for (auto v: tmp) { cerr << " " << v; } cerr << endl; } lint total = 0; for (int i = 0; i < (int) tmp.size() - 1; ++i) { lint diff = tmp[i + 1] - tmp[i]; total += diff * (diff - 1) / 2; total %= MOD; } if (DEBUG) { cerr << "return: " << total << endl; } return total; } int main(void) { int n; cin >> n; vector<int> a(n), b(n); for (int i = 0; i < n; ++i) { cin >> a[i]; a[i]--; } for (int i = 0; i < n; ++i) { cin >> b[i]; b[i]--; } vector<int> ainv = inverse(a), binv = inverse(b); vector<int> acycle, aperiod, aorder; vector<int> bcycle, bperiod, border; cycles(ainv, acycle, aperiod, aorder); cycles(binv, bcycle, bperiod, border); if (DEBUG) { cerr << "acycle: "; for (int i = 0; i < n; ++i) { cerr << " " << acycle[i]; } cerr << endl; cerr << "bcycle: "; for (int i = 0; i < n; ++i) { cerr << " " << bcycle[i]; } cerr << endl; } map<pii, vector<int> > data; for (int i = 0; i < n; ++i) { pii now(acycle[i], bcycle[i]); data[now].push_back(i); } if (DEBUG) { for (auto value: data) { cerr << "data "; cerr << value.first.first << " " << value.first.second << ":"; for (auto p: value.second) { cerr << " " << p; } cerr << endl; } } lint total = 0; for (map<pii, vector<int> >::iterator it = data.begin(); it != data.end(); ++it) { pii cycles = it->first; vector<int> indices = it->second; int acy = cycles.first; int bcy = cycles.second; int p_gcd = gcd(aperiod[acy], bperiod[bcy]); if (DEBUG) { cerr << "acy, bcy = " << acy << " " << bcy << ", pgcd = " << p_gcd << endl; } vector<vector<pii> > data(p_gcd); for (int i = 0; i < (int) indices.size(); ++i) { int index = indices[i]; int aord = aorder[index]; int bord = border[index]; int diff = (aord - bord) % p_gcd; if (diff < 0) { diff += p_gcd; } data[diff].push_back(pii(aord, bord)); } for (int i = 0; i < p_gcd; ++i) { total += doit(data[i], aperiod[acy], bperiod[bcy], p_gcd, i); total %= MOD; } } #if 0 for (int i = 0; i < n; ++i) { // (i, i) no gyaku lint count = 0; int ai = i; int bi = i; while (true) { if (count > 0 && ai == bi) { break; } ai = ainv[ai]; bi = binv[bi]; count++; } total = (total + count * (count - 1) / 2) % MOD; } #endif cout << total << endl; }