Submission #61422

#	Time	Username	Problem	Language	Result	Execution time	Memory
61422		Benq	Asceticism (JOI18_asceticism)	C++11	100 / 100	58 ms	1536 KiB

asceticism

#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 = 100005;

template<int SZ> struct Combo {
    int fac[SZ+1], ifac[SZ+1];
    
    ll mul(ll a, ll b) { return a*b%MOD; }
    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); }

    Combo() {
        fac[0] = ifac[0] = 1;
    	FOR(i,1,SZ+1) 
    	    fac[i] = mul(i,fac[i-1]), ifac[i] = inv(fac[i]);
    }
    
    ll comb(ll a, ll b) {
        if (a < b || b < 0) return 0;
        return mul(mul(fac[a],ifac[b]),ifac[a-b]);
    }
};

Combo<MX> C;

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 N,K;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K; K --;
    ll ans = 0;
    F0R(i,K+1) {
        ll pro = mul(po(K+1-i,N),C.comb(N+1,i));
        if (i&1) ans = sub(ans,pro);
        else ans = ad(ans,pro);
        // ans += (-1)^i*(K+1-i)^N*(N+1)Ci
    }
    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
*/

• Subtask 1: 4 / 4
• Subtask 2: 20 / 20
• Subtask 3: 25 / 25
• Subtask 4: 51 / 51