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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
ll po (ll b, ll p) { return !p?1:po(b*b%MOD,p/2)*(p&1?b:1)%MOD; }
ll inv (ll b) { return po(b,MOD-2); }
ll ad(ll a, ll b) { return (a+b)%MOD; }
ll sub(ll a, ll b) { return (a-b+MOD)%MOD; }
ll mul(ll a, ll b) { return a*b%MOD; }
ll divi(ll a, ll b) { return mul(a,inv(b)); }
int N, dp[1000][501], in[1001];
vi rm;
map<int,int> m;
vpi v;
void input() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N;
F0R(i,N) {
int a,b; cin >> a >> b;
v.pb({a,b});
m[a-1] = m[b] = 0;
}
FOR(i,1,1001) in[i] = inv(i);
int co = 0;
for (auto& a: m) {
rm.pb(a.f);
a.s = co++;
}
}
int main() {
input();
F0R(i,sz(v)) {
auto a = v[i];
int l = m[a.f-1]+1, r = m[a.s];
int csum = 1;
FOR(j,1,r+1) {
int tmp = 0; F0R(k,i) tmp = ad(tmp,dp[j][k]);
if (l <= j) {
int w = rm[j]-rm[j-1];
FORd(k,1,i+1) dp[j][k] = ad(dp[j][k],mul(dp[j][k-1],mul(w-k,in[k+1])));
dp[j][0] = ad(dp[j][0],mul(w,csum));
}
csum = ad(csum,tmp);
}
}
int ans = 0;
FOR(i,1,sz(m)) F0R(j,sz(v)) ans = ad(ans,dp[i][j]);
cout << ans;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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