[USACO 2013 Nov G]Empty Stalls
题号:NC24414
时间限制:C/C++/Rust/Pascal 1秒,其他语言2秒
空间限制:C/C++/Rust/Pascal 32 M,其他语言64 M
64bit IO Format: %lld

题目描述

Farmer John's new barn consists of a huge circle of N stalls (2 <= N <= 3,000,000), numbered 0..N-1, with stall N-1 being adjacent to stall 0. 
 At the end of each day, FJ's cows arrive back at the barn one by one, each with a preferred stall they would like to occupy. However, if a cow's preferred stall is already occupied by another cow, she scans forward sequentially from this stall until she finds the first unoccupied stall, which she then claims. If she scans past stall N-1, she continues scanning from stall 0. 
Given the preferred stall of each cow, please determine the smallest index of a stall that remains unoccupied after all the cows have returned to the barn. Notice that the answer to this question does not depend on the order in which the cows return to the barn. 
In order to avoid issues with reading huge amounts of input, the input to this problem is specified in a concise format using K lines (1 <= K <= 10,000) each of the form: 
X Y A B 
One of these lines specifies the preferred stall for XY total cows: X cows prefer each of the stalls f(1) .. f(Y), where f(i) = (Ai + B) mod N. The values of A and B lie in the range 0...1,000,000,000.  
Do not forget the standard memory limit of 64MB for all problems.

输入描述:

* Line 1: Two space-separated integers: N and K.

* Lines 2..1+K: Each line contains integers X Y A B, interpreted as
above. The total number of cows specified by all these lines
will be at most N-1. Cows can be added to the same stall by
several of these lines.

输出描述:

* Line 1: The minimum index of an unoccupied stall.
示例1

输入

复制
10 3
3 2 2 4
2 1 0 1
1 1 1 7 

输出

复制
5

说明

INPUT DETAILS:
There are 10 stalls in the barn, numbered 0..9. The second line of input
states that 3 cows prefer stall (2*1+4) mod 10 = 6, and 3 cows prefer stall
(2*2+4) mod 10 = 8. The third line states that 2 cows prefer stall (0*1+1)
mod 10 = 1. Line four specifies that 1 cow prefers stall (1*1+7) mod 10 =
8 (so a total of 4 cows prefer this stall).

OUTPUT DETAILS:
All stalls will end up occupied except stall 5.