제출 #20149

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
20149		emppu	능력 (kriii4_S)	C++14	100 / 100	596 ms	1856 KiB

S

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>

#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <tuple>
#include <queue>
#include <deque>
#include <list>

#define f0(_X,_Y) for(int (_X)=0;(_X)<(_Y);++(_X))
#define f1(_X,_Y) for(int (_X)=1;(_X)<=(_Y);++(_X))
#define ff(_X,_Y,_Z) for(int (_X)=(_Y);(_X)<=(_Z);++(_X))
#define fF(_X,_Y,_Z) for(int (_X)=(_Y);(_X)<(_Z);++(_X))
#define rf0(_X,_Y) for(int _X=(_Y)-1;(_X)>=0;--(_X))
#define rf1(_X,_Y) for(int _X=(_Y);(_X)>0;--(_X))
#define rff(_X,_Y,_Z) for(int _X=(_Y);(_X)>=(_Z);--(_X))
#define rfF(_X,_Y,_Z) for(int _X=(_Y);(_X)>(_Z);--(_X))

#define PRT(_X) cout<< #_X ": "<<_X<<endl;
#define TIME fprintf(stderr,"time : %.2f sec\n",double(clock())/CLOCKS_PER_SEC)
#define FIN freopen("input","r",stdin)

#define scan1(_X) scanf("%d",&_X);
#define scan2(_X,_Y) scanf("%d%d",&_X,&_Y);
#define scan3(_X,_Y,_Z) scanf("%d%d%d",&_X,&_Y,&_Z);
#define define1(_1) int _1; scan1(_1)
#define define2(_1,_2) int _1,_2; scan2(_1,_2)
#define define3(_1,_2,_3) int _1,_2,_3; scan3(_1,_2,_3)
#define EXPAND(_1) _1
#define SELECT(_1,_2,_3,_4,NAME,...) NAME
#define scan(...) EXPAND(SELECT(__VA_ARGS__, scan4, scan3, scan2, scan1)(__VA_ARGS__))
#define define(...) EXPAND(SELECT(__VA_ARGS__, define4, define3, define2, define1)(__VA_ARGS__))
#define print(_X) printf("%d\n",_X)
#define PAIR_STRUCT(_T,_X,_Y,...) struct _T{int _X,_Y,##__VA_ARGS__; bool friend operator < (const _T &p, const _T &q){if(p._X!=q._X) return p._X<q._X; return p._Y<q._Y;}}

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

const int N = 5005;
int p[N],pR[N];
int d[N],d2[N];

const int MOD = 1000000007;
ll mul(ll a, ll b){return a*b%MOD;}
ll add(ll a, ll b){return (a+b)%MOD;}
ll inv(ll a, ll b=MOD)
{
	ll s = 0;  ll  old_s = 1;
	ll r = b;  ll  old_r = a;
	while (r!=0)
	{
		ll quotient = old_r / r;
		tie(old_r, r) = make_tuple(r, old_r - quotient * r);
		tie(old_s, s) = make_tuple(s, old_s - quotient * s);
	}
	ll inv = old_s%MOD;
	if(inv<0) inv += MOD;
	return inv;
}


int dv[N];
int deal[N];
int _inv[N];
int main()
{
	// divisors
	ff(i,2,5000)
	{
		for(int j=i;j<=5000;j+=i) if(!dv[j]) dv[j]=i;
	}
	// inverse
	_inv[1]=1;
	ff(i,2,5000)
	{
		if(dv[i]!=i)
			_inv[i]=  mul(_inv[dv[i]],_inv[i/dv[i]]);
		else
			_inv[i]= inv(i);
	}

	define(n);
	const int denominator = 1000000000;
	const ll denominator_inv = inv(denominator);
	f1(i,n) { scan(p[i],deal[i]); pR[i] = mul(denominator-p[i],denominator_inv); p[i]=mul(p[i],denominator_inv); }

	d[0]=1;
	f1(i,n) rf1(j,i) d[j] = add(d[j],mul(pR[i],d[j-1]));
	ll ans=0;
	f1(i,n)
	{
		ll nCm_inv = _inv[n];
		// 1/nCm = 1*2*..*m/(n*(n-1)..(n-m+1))
		ll dmg = 0;
		ff(m,0,n-1)
		{
			if(!m)
				d2[m]=1;
			else
			{
				d2[m] = add(d[m],MOD - mul(d2[m-1],pR[i]));
				nCm_inv = mul(mul(nCm_inv, m), _inv[n-m]);
			}
			dmg = add(dmg,mul(d2[m],nCm_inv));
		}
		ans = add(ans, mul(dmg,mul(p[i],deal[i])));
	}

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

	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 90 / 90