51Nod - 1667 概率好题

Description

懒得写题面了。

Solution

假设甲在第 \(i\) 个区间选的是 \(R_i-x_i\) ，乙在第 \(i\) 个区间选的是 \(L_i+y_i\) 。若甲胜利则有

\[\sum_{i=1}^{k_1}R_i-x_i>\sum_{i=1}^{k_2}L_i+y_i \]

其中

\[0\le x_i\le R_i-L_i,0\le y_i\le R_i-L_i \]

移一下项

\[\sum_{i=1}^{k_1}x_i+\sum_{i=1}^{k_2}y_i<\sum_{i=1}^{k_1}R_i-\sum_{i=1}^{k_2}L_i \]

令

\[m=\sum_{i=1}^{k_1}R_i-\sum_{i=1}^{k_2}L_i \]

则有

\[\sum_{i=1}^{k_1}x_i+\sum_{i=1}^{k_2}y_i\le m - 1 \]

定义新变量 \(k\in[0,+\infty]\) 则有

\[\sum_{i=1}^{k_1}x_i+\sum_{i=1}^{k_2}y_i+k= m - 1 \]

于是就有了 \(k_1+k_2+1\) 个变量。简单容斥一下就好了。平局的情况把 \(k\) 上届置为 \(0\) ，等式右边改成 \(m\) 就行了。

#include<bits/stdc++.h>
using namespace std;

template <class T> void read(T &x) {
	x = 0; bool flag = 0; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == 45) flag = 1;
	for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48; if (flag) x = -x;
}

#define N 30
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define P 1000000007
#define INF 0x3f3f3f3f
#define ll long long

int k1, k2, m, up[N];
ll ans;

ll qpow(ll x, int k = P - 2) {
	ll ret = 1;
	for (; k; k >>= 1, (x *= x) %= P) if (k & 1) (ret *= x) %= P;
	return ret;
}

ll C(int n, int m) {
	if (n < m) return 0;
	ll ret = 1;
	rep(i, n - m + 1, n) (ret *= i) %= P;
	rep(i, 1, m) (ret *= qpow(i)) %= P;
	return ret;
}

void dfs(int x, bool cur, int m) {
	if (m < 0) return;
	if (x == k1 + k2 + 1) {
		int t = (up[x] == 0);
		if (cur) ans = (ans + P - C(m + k1 + k2 - t, k1 + k2 - t)) % P;
		else (ans += C(m + k1 + k2 - t, k1 + k2 - t)) %= P;
		return;
	}
	dfs(x + 1, 1 - cur, m - up[x] - 1), dfs(x + 1, cur, m);
}

void calcInv() {
	rep(i, 1, k1 + k2) (ans *= qpow(up[i] + 1)) %= P;
}

int main() {
	int T; read(T);
	while (T--) {
		m = 0;
		read(k1);
		rep(i, 1, k1) {
			int l, r; read(l), read(r);
			up[i] = r - l, m += r;
		}
		read(k2);
		rep(i, 1, k2) {
			int l, r; read(l), read(r);
			up[i + k1] = r - l, m -= l;
		}
		up[k1 + k2 + 1] = INF, m--, ans = 0;
		dfs(1, 0, m), calcInv();
		ll t = ans; ans = 0;
		up[k1 + k2 + 1] = 0, m++;
		dfs(1, 0, m), calcInv();
		printf("%lld %lld %lld\n", t, ans, (1 + P + P - t - ans) % P);
	}
	return 0;
}