結果

問題 No.2120 場合の数の下8桁
ユーザー leaf_1415leaf_1415
提出日時 2022-11-04 21:40:47
言語 C++11
(gcc 13.3.0)
結果
MLE  
実行時間 -
コード長 3,781 bytes
コンパイル時間 978 ms
コンパイル使用メモリ 102,836 KB
実行使用メモリ 816,512 KB
最終ジャッジ日時 2024-07-18 19:23:47
合計ジャッジ時間 10,827 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
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 * 7 MLE * 1 -- * 12
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 <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#include <array>
#include <unordered_set>
#include <unordered_map>
#define rep(x, s, t) for(ll x = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(ll x = (s); (x) >= (t); (x)--)
#define reps(x, s) for(ll x = 0; (x) < (ll)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define outl(...) dump_func(__VA_ARGS__)
#define outf(x) cout << fixed << setprecision(16) << (x) << endl
#define pb push_back
#define fi first
#define se second
#define inf 2e18
#define eps 1e-9
const double PI = 3.1415926535897932384626433;
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
map<ll, ll> mp;
ll a[55], b[55], m[55];
bool prime[10000005];
ll gcd(ll a, ll b){
if(b == 0) return a;
return gcd(b, a%b);
}
//ax+by = gcd(a, b)(x, y)gcd(a, b)
ll extgcd(ll a, ll b, ll &x, ll &y)
{
if(b == 0){
x = 1, y = 0;
return a;
}
ll xx, yy;
ll d = extgcd(b, a%b, xx, yy);
x = yy, y = xx-(a/b)*yy;
return d;
}
//a^{-1} (mod m)(gcd(a, m)!=1)-1
ll mod_inverse(ll a, ll m)
{
ll x, y;
if(extgcd(a, m, x, y) != 1) return -1;
return (x%m + m) % m;
}
//ax = b (mod m)x(mod m/gcd(a, m))(b%gcd(a, m)!=0)(0, -1)
P congruence(ll a, ll b, ll m)
{
ll d = gcd(a, m);
if(b % d) return make_pair(0, -1);
a /= d, b /= d, m /= d;
return make_pair(b * mod_inverse(a, m) % m, m);
}
//a_i*x = b_i (mod m_i)(i = 1, 2, ..., n)(x, M)(0, -1)
P SimultaneousCongruence(ll a[], ll b[], ll m[], ll n)
{
ll x = 0, M = 1;
for(int i = 1; i <= n; i++){
P res = congruence(a[i]*M, (b[i]-a[i]*x%m[i]+m[i])%m[i], m[i]);
if(res.second == -1) return res;
x += M*res.first, M *= res.second;
}
return make_pair(x, M);
}
const int FACT_MAX = 10000005;
ll q[FACT_MAX], e[FACT_MAX];
ll modpow(ll a, ll n, ll mod)
{
if(n == 0) return 1;
if(n % 2){
return ((a%mod) * (modpow(a, n-1, mod)%mod)) % mod;
}
else{
return modpow((a*a)%mod, n/2, mod) % mod;
}
}
void make_fact(ll p, ll mod)
{
ll qval = 1, eval = 0;
q[0] = 1, e[0] = 0;
for(int i = 1; i < FACT_MAX; i++){
ll t = i;
while(t % p == 0){
eval++;
t /= p;
}
qval *= t, qval %= mod;
q[i] = qval, e[i] = eval;
}
}
ll comb(ll n, ll k, ll p, ll ex, ll mod)
{
ll eval = e[n] - e[k] - e[n-k];
if(eval >= ex) return 0;
ll ret = q[n] * mod_inverse(q[k]*q[n-k]%mod, mod) % mod;
ret *= modpow(p, eval, mod), ret %= mod;
return ret;
}
ll n, k;
ll calc(ll p, ll ex, ll mod)
{
make_fact(p, mod);
//mod = p^ex
return comb(n, k, p, ex, mod);
}
//M
int main(void)
{
cin >> n >> k;
ll M = 100000000;
for(int i = 2; i < 10000005; i++){
if(prime[i]) continue;
for(int j = 2*i; j < 10000005; j+=i) prime[j] = true;
}
for(int i = 2; i < 10000005; i++){
if(prime[i]) continue;
while(M%i == 0){
mp[i]++;
M /= i;
}
}
if(M > 1) mp[M]++;
ll id = 0;
for(auto it = mp.begin(); it != mp.end(); it++){
id++;
ll mod = 1;
for(int i = 0; i < it->second; i++) mod *= it->first;
a[id] = 1, b[id] = calc(it->first, it->second, mod), m[id] = mod;
}
printf("%08d\n", (int)SimultaneousCongruence(a, b, m, id).first);
return 0;
}
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