Submission #19922

#	Time	Username	Problem	Language	Result	Execution time	Memory
19922		tncks0121	순열 (kriii4_T)	C++14	100 / 100	97 ms	24636 KiB

T

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <deque>
#include <utility>
#include <bitset>
#include <limits.h>
#include <time.h>
#include <functional>
#include <numeric>
 
using namespace std;
typedef long long ll;
typedef unsigned long long llu;
typedef double lf;
typedef unsigned int uint;
typedef long double llf;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;

#define debug(format, ...) printf(format, __VA_ARGS__);


const ll MOD = (ll)1e9 + 7;

ll modpow (ll a, ll b) {
    a %= MOD;
    ll ret = 1;
    while(b > 0) {
        if(b & 1) ret = (ret * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return ret;
}

struct mint {
    ll val;
    mint(ll val = 0): val((val % MOD + MOD) % MOD) { }
    mint operator+(mint p) { return val + p.val; }
    mint operator-(mint p) { return val - p.val; }
    mint operator*(mint p) { return val * p.val; }
    mint operator/(mint p) { return val * modpow(p.val, MOD-2); }
};

int N, K;

const int N_ = 1005000;
mint inv[N_];
mint fac[N_], invfac[N_];


int main() {
	scanf("%d%d", &N, &K);

    // precalc section
    {
        inv[1] = 1;
        for(int i = 2; i < N_; i++) {
            inv[i] = inv[MOD % i] * -(MOD / i);
        }
      
        invfac[0] = fac[0] = 1;
        for(int i = 1; i < N_; i++) {
            fac[i] = fac[i-1] * i;
            invfac[i] = invfac[i-1] * inv[i];
        }
    }

    mint ans;
    for(int i = K+1; i <= N; i++) {
    	ans = ans + fac[N+1] * inv[i+1] * (N+1-i);
    }

    printf("%lld\n", ans.val);

    return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 92 / 92