結果

問題 No.2119 一般化百五減算
ユーザー ecotteaecottea
提出日時 2022-11-04 21:47:12
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 5,205 bytes
コンパイル時間 3,627 ms
コンパイル使用メモリ 236,924 KB
実行使用メモリ 8,448 KB
最終ジャッジ日時 2024-07-18 19:31:39
合計ジャッジ時間 4,698 ms
ジャッジサーバーID
(参考情報)
judge3 / judge2
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ファイルパターン 結果
other AC * 24 WA * 1
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ソースコード

diff #
プレゼンテーションモードにする

#ifndef HIDDEN_IN_VS //
//
#define _CRT_SECURE_NO_WARNINGS
//
#include <bits/stdc++.h>
using namespace std;
//
using ll = long long; // -2^63 2^63 = 9 * 10^18int -2^31 2^31 = 2 * 10^9
using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
using Graph = vvi;
//
const double PI = acos(-1);
const vi DX = { 1, 0, -1, 0 }; // 4
const vi DY = { 0, 1, 0, -1 };
int INF = 1001001001; ll INFL = 4004004004004004004LL;
double EPS = 1e-12;
//
struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp;
//
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), x))
#define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), x))
#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 n-1
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s t
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s t
#define repe(v, a) for(const auto& v : (a)) // a
#define repea(v, a) for(auto& v : (a)) // a
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a
#define smod(n, m) ((((n) % (m)) + (m)) % (m)) // mod
#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} //
#define EXIT(a) {cout << (a) << endl; exit(0);} //
//
template <class T> inline ll pow(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // true
    
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // true
    
//
template <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }
template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }
// Visual Studio
#ifdef _MSC_VER
#include "local.hpp"
// gcc
#else
inline int popcount(int n) { return __builtin_popcount(n); }
inline int popcount(ll n) { return __builtin_popcountll(n); }
inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : -1; }
inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : -1; }
inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; }
inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; }
#define gcd __gcd
#define dump(...)
#define dumpel(v)
#define dump_list(v)
#define input_from_file(f)
#define output_to_file(f)
#define Assert(b) { if (!(b)) while (1) cout << "OLE"; }
#endif
#endif //
//--------------AtCoder --------------
#include <atcoder/all>
using namespace atcoder;
//using mint = modint1000000007;
using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);
istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }
ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>;
//----------------------------------------
int main() {
// input_from_file("input.txt");
// output_to_file("output.txt");
ll n; int m;
cin >> n >> m;
// x = c[i] (mod b[i])
vl b(m), c(m);
rep(i, m) cin >> b[i] >> c[i];
rep(i, m) c[i] = smod(c[i], b[i]);
map<ll, ll, greater<ll>> b_to_c;
rep(i, m) {
if (b_to_c.count(b[i])) {
if (b_to_c[b[i]] != c[i]) EXIT("NaN");
}
else {
b_to_c[b[i]] = c[i];
}
}
dump(b_to_c);
ll x = 0, mod = 1;
repe(tmp, b_to_c) {
ll b, c;
tie(b, c) = tmp;
while (true) {
if (x % b == c) break;
if (x > n) EXIT("NaN");
x += mod;
}
mod = mod / gcd(b, mod) * b;
}
cout << x << endl;
}
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