标签:abc return 学生 重复 代码 题意 class htm cout
F : Four-tuples
输入
1 1 1 2 2 3 3 4 4
输出
1
给l1, r1, l2, r2, l3, r3, l4, r4? , 八个数据, 要求输出在区间[l1, r1] , [l2, r2] , [l3, r3] , [l4, r4?]
(记为A, B, C, D)范围内, 各取一个数(取作x1, x2, x3, x4), 并且x1 != x2, x2 != x3, x3 != x4, x4 != x1,
注意像x1 == x3是可以的
容斥原理
结果 = 每个区间值的个数相乘 - 四种相等的情况即x1=x2, x2=x3...... (前者好说, 关键是后面这部分)
四种相等的情况计算方法:(ABCD代表四个区间里的任意一个数)
#include <bits/stdc++.h> #include <algorithm> using namespace std; const int mod = 1e9 + 7; typedef long long ll; ll query2(ll l1, ll r1, ll l2, ll r2)//query求的是几个区间重复的数的个数, 以下也是 { ll l = max(l1, l2), r = min(r1, r2); return r-l+1>0? r-l+1 : 0; } ll query3(ll l1, ll r1, ll l2, ll r2, ll l3, ll r3) { ll l = max(l1, max(l2, l3)), r = min(r1, min(r2, r3)); return r-l+1>0? r-l+1 : 0; } ll query4(ll l1, ll r1, ll l2, ll r2, ll l3, ll r3, ll l4, ll r4) { ll l = max(max(l3, l4), max(l1, l2)), r = min(min(r3, r4), min(r1, r2)); return r-l+1>0? r-l+1 : 0; } int main() { int t; ll l1, l2, l3, l4, r1, r2, r3, r4; cin >> t; while( t --) { scanf("%lld%lld%lld%lld%lld%lld%lld%lld", &l1,&r1,&l2,&r2,&l3,&r3,&l4,&r4); ll len1 = r1 - l1 + 1; ll len2 = r2 - l2 + 1; ll len3 = r3 - l3 + 1; ll len4 = r4 - l4 + 1; ll sum = len1 * len2 % mod; sum = sum * len3 % mod; sum = sum * len4 % mod; ll ant = 0; ant = (ant + query2(l1, r1, l2, r2) % mod * len3 % mod * len4 % mod) % mod; ant = (ant + query2(l2, r2, l3, r3) % mod * len1 % mod * len4 % mod) % mod; ant = (ant + query2(l4, r4, l3, r3) % mod * len1 % mod * len2 % mod) % mod; ant = (ant + query2(l4, r4, l1, r1) % mod * len2 % mod * len3 % mod) % mod; // 第一步完成--------------------------------------------------------- ant = (ant - query3(l1, r1, l2, r2, l3, r3) % mod * len4 % mod + mod) % mod; ant = (ant - query3(l1, r1, l2, r2, l4, r4) % mod * len3 % mod + mod) % mod; ant = (ant - query3(l4, r4, l2, r2, l3, r3) % mod * len1 % mod + mod) % mod; ant = (ant - query3(l1, r1, l4, r4, l3, r3) % mod * len2 % mod + mod) % mod; // 第二步完成--------------------------------------------------------- ant = (ant - query2(l1, r1, l2, r2) % mod * query2(l4, r4, l3, r3) % mod + mod) % mod; ant = (ant - query2(l2, r2, l3, r3) % mod * query2(l4, r4, l1, r1) % mod + mod) % mod; // 第三步完成--------------------------------------------------------- ant = (ant + 3 * query4(l1, r1, l2, r2, l3, r3, l4, r4) % mod + mod) % mod; // 第四步完成--------------------------------------------------------- cout << (sum - ant + mod) % mod << endl; } return 0; }
“浪潮杯”第九届山东省ACM大学生程序设计竞赛 F: Four-tuples容斥定理
标签:abc return 学生 重复 代码 题意 class htm cout
原文地址:https://www.cnblogs.com/la-la-wanf/p/14628710.html