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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define int ll
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
#define INF INT_MAX
#define MOD 1000000007
#define all(x) x.begin(), x.end()
int binexp(int base, int pow, int M=MOD){
int res = 1;
while(pow){
if(pow & 1) res *= base;
pow >>= 1, base *= base;
if(M) base %= M, res %= M;
}
return res;
}
vi fac(5e2+10), inv_fac(5e2+10);
int nCk(int n, int k){
if(n-k < k) k = n-k;
int v = inv_fac[k];
while(k--){
v *= n; v %= MOD;
n--;
}
return v;
}
vii A; vi B; vvi dp;
int solve(int n, int c){
if(n == -1) return 1;
if(c == 0) return 1;
if(dp[n][c] != -1)
return dp[n][c];
auto &v = dp[n][c]; v = solve(n, c-1);
for(int x = n, cnt = 0; x >= 0; x--){
if(!(A[x].first <= B[c-1] and B[c] <= A[x].second))
continue;
for(int k = 0; k <= min(cnt, B[c]-B[c-1]-1); k++){
int val = nCk(B[c]-B[c-1], k+1)*nCk(cnt, k); val %= MOD;
val *= solve(x-1, c-1), val %= MOD;
v += val; v %= MOD;
}
cnt++;
}
return v;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
fac[0] = inv_fac[0] = 1;
for(int x = 1; x <= 5e2; x++)
fac[x] = (x*fac[x-1])%MOD,
inv_fac[x] = binexp(fac[x], MOD-2);
int N; cin >> N;
A.resize(N);
for(auto &[l, r] : A){
cin >> l >> r; r++;
B.push_back(l); B.push_back(r);
}
sort(all(B));
dp.resize(N, vi(B.size(), -1));
cout << solve(N-1, B.size()-1)-1 << endl;
return 0;
}
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