Submission #546286

#	Time	Username	Problem	Language	Result	Execution time	Memory
546286		zaneyu	Boat (APIO16_boat)	C++14	100 / 100	599 ms	2464 KiB

boat

/*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 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=len;
            for(int k=i-1;k>=0;k--){
                dp[i][j]+=1LL*dp[k][j-1]*c%MOD;
                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)];
        ans%=MOD;
    }
    cout<<ans<<'\n';
}

• Subtask 1: 9 / 9
• Subtask 2: 22 / 22
• Subtask 3: 27 / 27
• Subtask 4: 42 / 42