제출 #19245

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
19245		kaTkaHr	Σ (kriii4_P2)	C++14	100 / 100	3 ms	1428 KiB

P2

#include <stdio.h>
#include <algorithm>
#include <map>

using namespace std;

typedef long long ll;

const int MX = 105, MM = 1000000007;

ll rv(ll A){
	ll R = 1, B = MM-2;
	while(B){
		if( B&1 ) R = R * A % MM;
		A = A * A % MM; B /= 2;
	}
	return R;
}

struct frac{
	ll A, B;
	frac(ll A, ll B):A(A), B(B){}
	frac(){A = 0, B = 1;}
	frac operator+ (const frac &l)const{
		return frac((A * l.B + B * l.A) % MM, B * l.B % MM);
	}
	frac operator*(const frac &l)const{
		return frac(A*l.A % MM, B*l.B % MM);
	}
	frac operator/(const frac &l)const{
		return frac(A*l.B % MM, B*l.A % MM);
	}
	frac operator- (const frac &l)const{
		return frac((A*l.B - B*l.A%MM + MM) % MM, B*l.B % MM);
	}
	ll v(){ return A * rv(B) % MM; }
};

frac D[MX][MX], I[MX][MX];
int P, Q, N, K;

void print()
{
	for(int i = 1; i <= N; i++){
		for(int j = 1; j <= N; j++){
			printf("%lld,%lld ", D[i][j].A, D[i][j].B);
		}
		printf("\n");
	}
	printf("\n");
}

int main()
{
	int M;
	frac ans = frac(0, 1);
	scanf("%d", &M);
	for(int i = 1; i <= M; i++){
		int N, S;
		scanf("%d%d", &N, &S);
		ans = ans + frac(S, N);
	}
	printf("%lld\n", ans.v());
	/*
	scanf("%d%d%d%d", &P, &Q, &N, &K);
	for(int i = 1; i <= N; i++){
		D[i][i] = I[i][i] = frac(1, 1);
		if( i <= N-2 ) D[i][i+1] = frac(MM-Q, P);
		if( i >= 3) D[i][i-1] = frac(MM-P+Q, P);
	}
	print();
	for(int i = 1; i <= N; i++){
		for(int j = i+1; j <= N; j++){
			frac p = D[j][i] / D[i][i];
			for(int k = 1; k <= N; k++){
				D[j][k] = D[j][k] - D[i][k] * p;
				I[j][k] = I[j][k] - I[i][k] * p;
			}
		}
	}
	for(int i = N; i >= 1; i--){
		for(int j = i-1; j >= 1; j--){
			frac p = D[j][i] / D[i][i];
			for(int k = N; k >= 1; k--){
				D[j][k] = D[j][k] - D[i][k] * p;
				I[j][k] = I[j][k] - I[i][k] * p;
			}
		}
	}
	for(int i = 1; i <= N; i++){
		frac p = frac(1, 1) / D[i][i];
		for(int j = 1; j <= N; j++){
			D[i][j] = D[i][j] * p;
			I[i][j] = I[i][j] * p;
		}
	}
	print();

	for(int i = 1; i <= N; i++){
		for(int j = 1; j <= N; j++){
			printf("%lld ", I[i][j].v());
		}
		printf("\n");
	}
	printf("%lld", I[N][K].v());*/
}

• Subtask 1: 100 / 100