Submission #62266

#	Time	Username	Problem	Language	Result	Execution time	Memory
62266		Benq	The grade (info1cup18_thegrade)	C++11	100 / 100	93 ms	17312 KiB

thegrade

#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); }

int ad(int a, int b) { return (a+b)%MOD; }
int sub(int a, int b) { return (a-b+MOD)%MOD; }
int mul(int a, int b) { return (ll)a*b%MOD; }

int AD(int& a, int b) { return a = ad(a,b); }
int SUB(int& a, int b) { return a = sub(a,b); }
int MUL(int& a, int b) { return a = mul(a,b); }

int oc[1000001], num;
int fac[MX], ifac[MX], in[MX];
int q,p;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> q >> p;
    
    ll sum = p;
    int cdiv = 1;
    fac[0] = ifac[0] = 1;
    FOR(i,1,MX) {
        in[i] = inv(i);
        fac[i] = mul(fac[i-1],i);
        ifac[i] = mul(ifac[i-1],in[i]);
    }
    
    F0R(i,q) {
        int t,x; cin >> t >> x;
        if (t == 0) {
            num ++, sum -= x;
            MUL(cdiv,in[++oc[x]]);
        } else {
            num --, sum += x;
            MUL(cdiv,oc[x]--);
        }
        if (sum >= 0) cout << mul(mul(fac[num+sum],ifac[sum]),cdiv);
        else cout << -1;
        cout << "\n";
    }
    
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 15 / 15
• Subtask 6: 15 / 15
• Subtask 7: 20 / 20