第一条题解

#include <bits/stdc++.h>
#define FL(i, a, b) for (int i = (a); i <= (b); ++i)
#define FR(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long ll;
const int N = 5e3 + 10;
const int MOD = 1e9 + 7;
const ll INFLL = 1e18;
int n, a[N], b[N], c[N], ic[N];
int pl[N], pr[N], v[N], iv[N];
ll mi[2][N][N], ct[2][N][N];
ll s[N], t[N], f[N], g[N];
int Qpow(int a, int b) {
	int ret = 1;
	for (; b; b >>= 1) {
		if (b & 1)
			ret = (ll)ret * a % MOD;
		a = (ll)a * a % MOD;
	}
	return ret;
}
int Inv(int x) {
	return Qpow(x, MOD - 2);  
}
void Solve() {
	scanf("%d", &n);
	FL(i, 1, n) {
		scanf("%d", &a[i]);
		t[i] = t[i - 1] + (i & 1? -a[i] : a[i]);
	}
	FL(i, 1, n) {
		scanf("%d", &b[i]);
		s[i] = s[i - 1] + b[i];
	}
	v[0] = iv[0] = 1;
	FL(i, 1, n) {
		scanf("%d", &c[i]);
		v[i] = (ll)v[i - 1] * c[i] % MOD;
		ic[i] = Inv(c[i]);
		iv[i] = Inv(v[i]);
	}
	FL(i, 1, n) {
		mi[0][i][i - 1] = mi[1][i][i - 1] = INFLL;
		ct[0][i][i - 1] = ct[1][i][i - 1] = 0;
		FL(j, i, n) {
			FL(k, 0, 1) {
				mi[k][i][j] = mi[k][i][j - 1];
				ct[k][i][j] = ct[k][i][j - 1];
			}
			mi[j & 1][i][j] = min(mi[j & 1][i][j], (ll)b[j]);
			ct[j & 1][i][j] = (ct[j & 1][i][j] + ic[j]) % MOD;
		}
	}
	FL(i, 1, n) {
		int d = a[i];
		FL(j, i, n) {
			d = a[j + 1] - d;
			if (j == n || d <= 0) {
				pl[i] = j;
				break;
			}
		}
		d = a[i];
		FR(j, i, 1) {
			d = a[j - 1] - d;
			if (j == 1 || d <= 0) {
				pr[i] = (j > 1 && !d? j - 1 : j);
				break;
			}
		}
	}
	fill(f, f + n + 1, -INFLL);
	fill(g, g + n + 1, 0);
	f[0] = 0, g[0] = 1;
	FL(i, 0, n) {
		FL(j, i + 1, n) {
			int l = max(i + 1, pr[j]), r = min(j, pl[i + 1]);
			if (l > r){
				continue;
			}

			int h = (ll)g[i] * v[j] % MOD * iv[i] % MOD;
			if (t[j] - t[i] == 0) {
				f[j] = max(f[j], f[i] + (s[j] - s[i]));
				g[j] = (g[j] + h) % MOD;
			} else if (t[j] - t[i] > 0) {
				f[j] = max(f[j], f[i] + (s[j] - s[i]) - mi[0][l][r]);
				g[j] = (g[j] + (ll)h * ct[0][l][r]) % MOD;
			} else {
				f[j] = max(f[j], f[i] + (s[j] - s[i]) - mi[1][l][r]);
				g[j] = (g[j] + (ll)h * ct[1][l][r]) % MOD;
			}
		}
	}
	printf("%lld %lld\n", f[n], g[n]);
}
int main() {
	int T;
	scanf("%*d %d", &T);
	while (T--) {
		Solve();
	}
	return 0;
}