十二省NOI“省选”联考模测（第二场）A抽卡大赛

/*
dp维护整体的概率， 每次相当于回退一格然后重新dp一格 

*/
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#define ll long long
#define M 202
#define mmp make_pair
using namespace std;
int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
const int mod = 1000000007;
int poww(int a, int b)
{
	int ans = 1, tmp = a;
	for(; b; b >>= 1, tmp = 1ll * tmp * tmp % mod) if(b & 1) ans = 1ll * ans * tmp % mod;
	return ans;
}
void add(int &x, int y)
{
	x += y;
	x -= x >= mod ? mod : 0;
	x += x < 0 ? mod : 0;
}
struct Note
{
	int a, g, p;
	bool operator < (const Note &b) const
	{
		return this->a < b.a;
	}
}note[M][M], sta[M * M];
int m[M], q[M], n, v[M], tp;
int f[M], g[M], d[M], ans[M];
void work(int x)
{
	if(x == 0)
	{
		for(int i = 0; i <= n; i++) g[i] = g[i + 1];
		return;
	}
	int y = (1 + mod - x), invx = poww(x, mod - 2);
	memset(d, 0, sizeof(d));
	for(int i = 0; i <= n; i++)
	{
		d[i] = 1ll * g[i] * invx % mod;
		add(g[i + 1], -1ll * d[i] * y % mod);
	}
	for(int i = 0; i <= n; i++) g[i] = d[i];

}

int main()
{
	n = read();
	int inv = poww(100 ,mod - 2);
	for(int i = 1; i <= n; i++)
	{
		m[i] = read();
		for(int j = 1; j <= m[i]; j++)
		{
			note[i][j].a = read(), note[i][j].g = 1ll * (100 - read()) * inv % mod, note[i][j].p = read();
			add(q[i], note[i][j].p);
			sta[++tp] = (Note) {note[i][j].a, i, j};
		}
		int inv = poww(q[i], mod - 2);
		for(int j = 1; j <= m[i]; j++) note[i][j].p = 1ll * note[i][j].p * inv % mod;
	}
	sort(sta + 1, sta + tp + 1);
	for(int i = 1; i <= n; i++) v[i] = read();
	g[n] = 1;
	for(int now = 1; now <= tp; now++)
	{
		int i = sta[now].g, j = sta[now].p;
		work(f[i]);
		for(int a = 0; a <= n; a++) add(ans[i], 1ll * note[i][j].p * v[a + 1] % mod * g[a] % mod * note[i][j].g % mod);
		add(f[i], note[i][j].p);
		for(int a = n; a >= 0; a--)
		{
			 g[a] = 1ll * g[a] * f[i] % mod;
			 if(a) add(g[a], 1ll * (1 + mod - f[i]) * g[a - 1] % mod);
		}
	}
	for(int i = 1; i <= n; i++) cout << ans[i] << "\n";
	return 0;
}