이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define N 300005
using namespace std;
const int mod = 1e9 + 7;
long long fact[N];
long long ifact[N];
long long binpow(long long a,long long b){
long long ret = 1;
while(b){
if(b & 1)
ret = ret*a%mod;
a = a*a%mod;
b >>= 1;
}
return ret;
}
long long C(int n,int r){
return fact[n] * ifact[r]%mod*ifact[n-r]%mod;
}
long long val[N][5];
void solve(){
int n;
cin >> n;
map<int,int> cnt;
for(int i = 1;i<=n;i++){
int x;
cin >> x;
cnt[x]++;
}
vector<int> v;
for(auto u:cnt)
v.push_back(u.second);
if(cnt.size() == 2){
if(v[0] < 2 || v[1] < 3){
cout << 0;
return;
}
cout << C(v[0]-2+1,1) *C(v[1]-3+2,2) %mod;
return;
}
for(int i = 2;i<=n;i++){
val[i][1] = 1;
}
for(int i = 3;i<=n;i++){
val[i][2] = (binpow(2,i-1) - 2 + mod)%mod;
}
for(int i = 4;i<=n;i++){
//val[n][3] = (3^n - 2^(n+2) - 2n + 11)/4
val[i][3] = (binpow(3,i) - binpow(2,i+2) - 2*i + 11 + 2*mod)%mod*binpow(4,mod-2)%mod;
//cout << i << " " << val[i][3] << endl;
}
for(int i = 5;i<=n;i++){
for(int j = 2;j<=i-3;j++){
val[i][4] = (val[i][4] + C(i-1,j-1) *
( (binpow(3,i-j) - binpow(2,i-j+1) - 2 * (i-j) + 3 + 2*mod )%mod* binpow(4,mod-2)%mod )%mod);
}
val[i][4] = val[i][4] *2 %mod;
//cout << i << " " << val[i][4] << endl;
}
cout << val[n][4] << endl;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
#ifdef Local
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
fact[0] = ifact[0]= 1;
for(int i = 1;i<N;i++){
fact[i] = fact[i-1] * i %mod;
ifact[i] = binpow(fact[i],mod-2);
}
int t = 1;
//cin >> t;
while(t--){
solve();
}
#ifdef Local
cout << endl << fixed << setprecision(2) << 1000.0*clock()/CLOCKS_PER_SEC << " milliseconds.";
#endif
}
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