結果

問題 No.187 中華風 (Hard)
ユーザー mdj982mdj982
提出日時 2019-04-23 02:24:11
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,633 bytes
コンパイル時間 1,682 ms
コンパイル使用メモリ 170,724 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-10-12 10:52:13
合計ジャッジ時間 2,608 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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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
5,248 KB
testcase_01 AC 2 ms
5,248 KB
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 AC 2 ms
5,248 KB
testcase_18 AC 2 ms
5,248 KB
testcase_19 AC 2 ms
5,248 KB
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
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ソースコード

diff #

#include <bits/stdc++.h>
using namespace std;

#pragma warning(disable : 4267) // "int n = (unsigned)size"

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ll = long long int;
using vll = vector<ll>; using vvll = vector<vll>; using vvvll = vector<vvll>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;

const double eps = 1e-9;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
#define Loop(i, n) for(int i = 0; i < int(n); i++)
#define Loopll(i, n) for(ll i = 0; i < ll(n); i++)
#define Loop1(i, n) for(int i = 1; i <= int(n); i++)
#define Loopll1(i, n) for(ll i = 1; i <= ll(n); i++)
#define Loopr(i, n) for(int i = int(n) - 1; i >= 0; i--)
#define Looprll(i, n) for(ll i = ll(n) - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = int(n); i >= 1; i--)
#define Looprll1(i, n) for(ll i = ll(n); i >= 1; i--)
#define Foreach(buf, container) for(auto buf : container)
#define Loopdiag(i, j, h, w, sum) for(int i = ((sum) >= (h) ? (h) - 1 : (sum)), j = (sum) - i; i >= 0 && j < (w); i--, j++)
#define Loopdiagr(i, j, h, w, sum) for(int j = ((sum) >= (w) ? (w) - 1 : (sum)), i = (sum) - j; j >= 0 && i < (h); j--, i++)
#define Loopdiagsym(i, j, h, w, gap) for (int i = ((gap) >= 0 ? (gap) : 0), j = i - (gap); i < (h) && j < (w); i++, j++)
#define Loopdiagsymr(i, j, h, w, gap) for (int i = ((gap) > (h) - (w) - 1 ? (h) - 1 : (w) - 1 + (gap)), j = i - (gap); i >= 0 && j >= 0; i--, j--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define printv(vector) Loop(ex_i, vector.size()) { cout << vector[ex_i] << " "; } cout << endl;
#define printmx(matrix) Loop(ex_i, matrix.size()) { Loop(ex_j, matrix[ex_i].size()) { cout << matrix[ex_i][ex_j] << " "; } cout << endl; }
#define quickio() ios::sync_with_stdio(false); cin.tie(0);
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
#define Comp(type_t) bool operator<(const type_t &another) const
#define fst first
#define snd second
bool nearlyeq(double x, double y) { return abs(x - y) < eps; }
bool inrange(ll x, ll t) { return x >= 0 && x < t; }
bool inrange(vll xs, ll t) { Foreach(x, xs) if (!(x >= 0 && x < t)) return false; return true; }
int ceillog2(ll x) { int ret = 0;	x--; while (x > 0) { ret++; x >>= 1; } return ret; }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (m * m <= x ? 0 : -1); }
ll ceilsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }
ll lcm(ll m, ll n) { return m * n / gcd(m, n); }

/*******************************************************/

// mx + ny = gcd(m, n), runtime error for (m, n) = (0, 0)
ll ex_euclid(ll m, ll n, ll &x, ll &y) {
	if (n == 0) { x = 1; y = 0; return m; }
	ll g = ex_euclid(n, m % n, y, x);
	y -= m / n * x;
	return g;
}

ll chinese_remainder_theorem(const vll &ps, const vll &rs, ll MOD) {
	ll p = 1, r = 0;
	Loop(i, ps.size()) {
		ll x, y;
		ll g = ex_euclid(p, ps[i], x, y);
		ll z = rs[i] - r;
		if (z % g != 0) return -1;
		ll d = z / g;
		r = (x * d * p + r) % MOD;
		p = (p * ps[i] / g) % MOD;
	}
	return r;
}

int main() {
	int n; cin >> n;
	vll ps(n), rs(n);
	Loop(i, n) {
		cin >> rs[i] >> ps[i];
	}
	cout << chinese_remainder_theorem(ps, rs, ll(1e9) + 7) << endl;
}
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