问题 F: Four-tuples

问题 F: Four-tuples

时间限制: 10 Sec   内存限制: 128 MB

题目描述

Given l1,r1,l2,r2,l3,r3,l4,r4, please count the number of four-tuples (x1,x2,x3,x4) such that li≤ xi≤ ri and x1≠x2,x2≠x3,x3≠x4,x4≠x1. The answer should modulo 10^9+7 before output.

输入

The input consists of several test cases. The first line gives the number of test cases, T(1≤ T≤ 10^6).
For each test case, the input contains one line with 8 integers l1,r1,l2, r2, l3,r3,l4,r4(1≤ li≤ ri≤ 10^9)

输出

For each test case, output one line containing one integer, representing the answer.

样例输入

1

1 1 2 2 3 3 4 4

样例输出

1

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
int main() {
	//std::ios::sync_with_stdio(false);
	int t;
	//cin >> t;
	scanf ("%d", &t);
	while(t--) {
		ll l1, r1, l2, r2, l3, r3, l4, r4;
	//	cin >> l1 >> r1 >> l2 >> r2 >> l3 >> r3 >> l4 >> r4;
		scanf ("%lld %lld %lld %lld %lld %lld %lld %lld", &l1, &r1, &l2, &r2, &l3, &r3, &l4, &r4);
		ll t1 = r1 - l1 + 1;
		ll t2 = r2 - l2 + 1;
		ll t3 = r3 - l3 + 1;
		ll t4 = r4 - l4 + 1;
		ll res = t1 * t2 % mod * t3 % mod * t4 % mod;
		ll t12 = max(min(r1, r2) - max(l1, l2) + 1ll, 0ll);
		ll t23 = max(min(r2, r3) - max(l2, l3) + 1ll, 0ll);
		ll t34 = max(min(r3, r4) - max(l3, l4) + 1ll, 0ll);
		ll t41 = max(min(r1, r4) - max(l1, l4) + 1ll, 0ll);
		ll t123 = max(min(r1, min(r2, r3)) - max(l1, max(l2, l3)) + 1ll, 0ll);
		ll t124 = max(min(r1, min(r2, r4)) - max(l1, max(l2, l4)) + 1ll, 0ll);
		ll t423 = max(min(r4, min(r2, r3)) - max(l4, max(l2, l3)) + 1ll, 0ll);
		ll t143 = max(min(r1, min(r4, r3)) - max(l1, max(l4, l3)) + 1ll, 0ll);
		ll t1234 = max(min(min(r1, r2), min(r3, r4)) - max(max(l1, l2), max(l3, l4)) + 1ll, 0ll);
		ll temp1 = (t12 * t3 % mod * t4 % mod + t23 * t1 % mod * t4 % mod + t34 * t1 % mod * t2 % mod + t41 * t2 % mod * t3 % mod ) % mod;
		ll temp2 = (t123 * t4 % mod + t124 * t3 % mod + t423 * t1 % mod + t143 * t2 % mod + t12 * t34 % mod + t23 * t41 % mod) % mod;
		res = (res % mod - temp1 % mod + mod) % mod;
		res =(res % mod + temp2 % mod) % mod;
		res = (res % mod - 3 * t1234 % mod + mod) % mod;
	//	cout << res << endl;
		printf ("%lld\n", res);
	}
	return 0;
}