結果

問題 No.186 中華風 (Easy)
ユーザー anta
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
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 23
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#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 0x3f3f3f3f3f3f3f3fLL
using 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;
}
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