結果
問題 | No.186 中華風 (Easy) |
ユーザー |
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提出日時 | 2015-04-19 23:41:05 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 3 ms / 2,000 ms |
コード長 | 3,388 bytes |
コンパイル時間 | 1,061 ms |
コンパイル使用メモリ | 101,720 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-19 18:22:40 |
合計ジャッジ時間 | 1,824 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 23 |
ソースコード
#include <string>#include <vector>#include <algorithm>#include <numeric>#include <set>#include <map>#include <queue>#include <iostream>#include <sstream>#include <cstdio>#include <cmath>#include <ctime>#include <cstring>#include <cctype>#include <cassert>#include <limits>#include <functional>#include <bitset>#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))#if defined(_MSC_VER) || __cplusplus > 199711L#define aut(r,v) auto r = (v)#else#define aut(r,v) __typeof(v) r = (v)#endif#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)#define all(o) (o).begin(), (o).end()#define pb(x) push_back(x)#define mp(x,y) make_pair((x),(y))#define mset(m,v) memset(m,v,sizeof(m))#define INF 0x3f3f3f3f#define INFL 0x3f3f3f3f3f3f3f3fLLusing namespace std;typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }template<typename T>T gcd(T x, T y){if(y==0)return x;else return gcd(y,x%y);}template<typename T>T lcm(T x, T y){ return x == 0 ? 0 : x/gcd(x,y)*y; }vector<bool> isprime;vector<int> primes;void sieve(int n){if((int)isprime.size() >= n+1) return;isprime.assign(n+1, true);isprime[0] = isprime[1] = false;int sqrtn = (int)(sqrt(n * 1.) + .5);for(int i = 2; i <= sqrtn; i ++) if(isprime[i]) {for(int j = i * i; j <= n; j += i)isprime[j] = false;}primes.clear();for(int i = 2; i <= n; i ++) if(isprime[i])primes.push_back(i);}typedef int FactorsInt;typedef vector<pair<FactorsInt,int> > Factors;void primeFactors(FactorsInt x, Factors &out_v) {out_v.clear();int sqrtx = (int)(sqrt(x*1.) + 10.5);sieve(sqrtx);for(vector<int>::const_iterator p = primes.begin(); p != primes.end(); ++ p) {if(*p > sqrtx) break;if(x % *p == 0) {int t = 1;x /= *p;while(x % *p == 0) {t ++;x /= *p;}out_v.push_back(make_pair(*p, t));}}if(x != 1) out_v.push_back(make_pair(x, 1));}long long inverse(signed long long a, const long long MOD) {a %= MOD;if(a < 0) a += MOD;signed long long b = MOD, u = 1, v = 0;while(b) {signed long long t = a / b;a -= t * b; swap(a, b);u -= t * v; swap(u, v);}if(u < 0) u += MOD;return u;}ll crt(ll a1, ll a2, ll n1, ll n2) {ll b = n1 % n2;ll t = inverse(b, n2);ll h = ((a2 - a1) * t % n2 + n2) % n2;return a1 + n1 * h;}int powpq(int p, int q) {int r = 1;rep(i, q) r *= p;return r;}int main() {sieve(31623);int N = 3;// scanf("%d", &N);Factors fs;map<int,pii> mods;bool ok = true;rep(i, N) {int X, Y;cin >> X >> Y;primeFactors(Y, fs);each(j, fs) {int p = j->first, q = j->second;int pq = powpq(p, q);pii &t = mods[p];int u = min(pq, powpq(p, t.first));ok &= t.second % u == X % u;if(t.first < q) {t.first = q;t.second = X % pq;}}}if(!ok) {puts("-1");return 0;}ll a = 0, n = 1;each(i, mods) {int p = i->first, q = i->second.first, x = i->second.second;int pq = powpq(p, q);a = crt(a, x, n, pq);n *= pq;}if(a == 0)a += n;printf("%lld\n", a);return 0;}