結果
問題 | No.2119 一般化百五減算 |
ユーザー | siganai |
提出日時 | 2022-11-04 22:10:59 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,488 bytes |
コンパイル時間 | 1,867 ms |
コンパイル使用メモリ | 209,960 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-07-18 19:59:31 |
合計ジャッジ時間 | 2,741 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 | 1 ms
5,248 KB |
testcase_01 | AC | 1 ms
5,376 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | WA | - |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 2 ms
5,376 KB |
testcase_06 | WA | - |
testcase_07 | AC | 2 ms
5,376 KB |
testcase_08 | WA | - |
testcase_09 | AC | 2 ms
5,376 KB |
testcase_10 | WA | - |
testcase_11 | AC | 2 ms
5,376 KB |
testcase_12 | AC | 2 ms
5,376 KB |
testcase_13 | WA | - |
testcase_14 | WA | - |
testcase_15 | WA | - |
testcase_16 | WA | - |
testcase_17 | WA | - |
testcase_18 | WA | - |
testcase_19 | WA | - |
testcase_20 | WA | - |
testcase_21 | AC | 14 ms
5,376 KB |
testcase_22 | WA | - |
testcase_23 | WA | - |
testcase_24 | WA | - |
ソースコード
//#pragma GCC target("avx") //#pragma GCC optimize("O3") //#pragma GCC optimize("unroll-loops") #include <bits/stdc++.h> #ifdef LOCAL #include <debug.hpp> #define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__) #else #define debug(...) (static_cast<void>(0)) #endif using namespace std; using ll = long long; using ld = long double; using pll = pair<ll,ll>; using pii = pair<int,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 vpii = vector<pii>; using vpll = vector<pll>; using vs = vector<string>; template<class T> using pq = priority_queue<T,vector<T>,greater<T>>; #define overload4(_1, _2, _3, _4, name, ...) name #define overload3(a,b,c,name,...) name #define rep1(n) for (ll UNUSED_NUMBER = 0; UNUSED_NUMBER < (n); ++UNUSED_NUMBER) #define rep2(i, n) for (ll i = 0; i < (n); ++i) #define rep3(i, a, b) for (ll i = (a); i < (b); ++i) #define rep4(i, a, b, c) for (ll i = (a); i < (b); i += (c)) #define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__) #define rrep1(n) for(ll i = (n) - 1;i >= 0;i--) #define rrep2(i,n) for(ll i = (n) - 1;i >= 0;i--) #define rrep3(i,a,b) for(ll i = (b) - 1;i >= (a);i--) #define rrep4(i,a,b,c) for(ll i = (a) + ((b)-(a)-1) / (c) * (c);i >= (a);i -= c) #define rrep(...) overload4(__VA_ARGS__, rrep4, rrep3, rrep2, rrep1)(__VA_ARGS__) #define all1(i) begin(i),end(i) #define all2(i,a) begin(i),begin(i)+a #define all3(i,a,b) begin(i)+a,begin(i)+b #define all(...) overload3(__VA_ARGS__, all3, all2, all1)(__VA_ARGS__) #define sum(...) accumulate(all(__VA_ARGS__),0LL) template<class T> bool chmin(T &a, const T &b){ if(a > b){ a = b; return 1; } else return 0; } template<class T> bool chmax(T &a, const T &b){ if(a < b){ a = b; return 1; } else return 0; } template<class T> auto min(const T& a){ return *min_element(all(a)); } template<class T> auto max(const T& a){ return *max_element(all(a)); } template<class... Ts> void in(Ts&... t); #define INT(...) int __VA_ARGS__; in(__VA_ARGS__) #define LL(...) ll __VA_ARGS__; in(__VA_ARGS__) #define STR(...) string __VA_ARGS__; in(__VA_ARGS__) #define CHR(...) char __VA_ARGS__; in(__VA_ARGS__) #define DBL(...) double __VA_ARGS__; in(__VA_ARGS__) #define LD(...) ld __VA_ARGS__; in(__VA_ARGS__) #define VEC(type, name, size) vector<type> name(size); in(name) #define VV(type, name, h, w) vector<vector<type>> name(h, vector<type>(w)); in(name) ll intpow(ll a, ll b){ ll ans = 1; while(b){if(b & 1) ans *= a; a *= a; b /= 2;} return ans;} ll modpow(ll a, ll b, ll p){ ll ans = 1; a %= p;while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; } ll GCD(ll a,ll b) { if(a == 0 || b == 0) return a + b; if(a % b == 0) return b; else return GCD(b,a%b);} ll LCM(ll a,ll b) { if(a == 0) return b; if(b == 0) return a;return a / GCD(a,b) * b;} namespace IO{ #define VOID(a) decltype(void(a)) struct setting{ setting(){cin.tie(nullptr); ios::sync_with_stdio(false);fixed(cout); cout.precision(12);}} setting; template<int I> struct P : P<I-1>{}; template<> struct P<0>{}; template<class T> void i(T& t){ i(t, P<3>{}); } void i(vector<bool>::reference t, P<3>){ int a; i(a); t = a; } template<class T> auto i(T& t, P<2>) -> VOID(cin >> t){ cin >> t; } template<class T> auto i(T& t, P<1>) -> VOID(begin(t)){ for(auto&& x : t) i(x); } template<class T, size_t... idx> void ituple(T& t, index_sequence<idx...>){ in(get<idx>(t)...);} template<class T> auto i(T& t, P<0>) -> VOID(tuple_size<T>{}){ ituple(t, make_index_sequence<tuple_size<T>::value>{});} #undef VOID } #define unpack(a) (void)initializer_list<int>{(a, 0)...} template<class... Ts> void in(Ts&... t){ unpack(IO :: i(t)); } #undef unpack constexpr int mod = 1000000007; //constexpr int mod = 998244353; static const double PI = 3.1415926535897932; template <class F> struct REC { F f; REC(F &&f_) : f(forward<F>(f_)) {} template <class... Args> auto operator()(Args &&...args) const { return f(*this, forward<Args>(args)...); }}; inline long long MOD(long long a, long long m) { return (a % m + m) % m; } long long extGcd(long long a, long long b, long long &p, long long &q) { if (b == 0) { p = 1; q = 0; return a; } long long d = extGcd(b, a%b, q, p); q -= a/b * p; return d; } // 中国剰余定理 // リターン値を (r, m) とすると解は x ≡ r (mod. m) // 解なしの場合は (0, -1) をリターン tuple<long long, long long,long long> ChineseRem(const vector<long long> &b, const vector<long long> &m,int N) { long long r = 0, M = 1; for (int i = 0; i < (int)b.size(); ++i) { long long p, q; long long d = extGcd(M, m[i], p, q); // p is inv of M/d (mod. m[i]/d) if ((b[i] - r) % d != 0) return make_tuple(0,-1,-1); long long tmp = (b[i] - r) / d * p % (m[i]/d); r += M * tmp; M *= m[i]/d; if(M > N) return make_tuple(MOD(r,M),M,-1); } return make_tuple(MOD(r, M),M,1); } int main() { INT(n,m); vl b(m),c(m); rep(i,m) cin >> b[i] >> c[i]; rep(i,m) { c[i] %= b[i]; if(c[i] < 0) c[i] += b[i]; } debug(b,c); auto ans = ChineseRem(c,b,n); if(get<1>(ans) == -1) cout << "NaN" << '\n'; else if(get<2>(ans) == 1) cout << get<0>(ans) << '\n'; else { int tmp = get<0>(ans); rep(i,m) { if((tmp - c[i]) % b[i]) cout << "NaN" << '\n'; return 0; } cout << tmp << '\n'; } }