/*input
2
1 1000000000
1 1000000000
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("O2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,ll>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll maxn=1e3+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
out<<P.f<<' '<<P.s;
return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?"\n":"");
return out;
}
ll mult(ll a,ll b){
return a*b%MOD;
}
ll mypow(ll a,ll b){
a%=MOD;
if(a==0) return 0;
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1) res=(res*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return res;
}
int l[maxn],r[maxn];
int dp[maxn][maxn];
int c[maxn][maxn];
int inv[maxn];
int32_t main(){
ios::sync_with_stdio(false),cin.tie(0);
int n;
cin>>n;
vector<int> v;
REP(i,n+2) inv[i]=mypow(i,MOD-2);
REP1(i,n) cin>>l[i]>>r[i],++r[i],v.pb(l[i]),v.pb(r[i]);
SORT_UNIQUE(v);
REP1(i,n) l[i]=lower_bound(ALL(v),l[i])-v.begin()+1,r[i]=lower_bound(ALL(v),r[i])-v.begin();
REP(i,sz(v)) dp[0][i]=1;
int ans=0;
REP1(i,n){
for(int j=l[i];j<=r[i];j++){
int cnt=0;
int len=v[j]-v[j-1];
int c=1;
for(int k=i-1;k>=0;k--){
dp[i][j]+=1LL*dp[k][j-1]*c;
dp[i][j]%=MOD;
if(l[k]<=j and j<=r[k]){
++cnt;
c=1LL*c*(len+cnt)%MOD*inv[cnt+1]%MOD;
}
}
}
REP1(j,sz(v)) dp[i][j]+=dp[i][j-1],dp[i][j]%=MOD;
ans+=dp[i][sz(v)];
}
cout<<ans<<'\n';
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
6 ms |
2260 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
6 ms |
2260 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
3 ms |
724 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
6 ms |
2260 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |