Links: https://ac.nowcoder.com/acm/contest/1086/F
Source: Cattle-off network
Title Description
A full moon casts some sort of spell on the cows and, like their cousins the wolves and coyotes, they bay at the moon -- mooing instead of howling, of course.
Each 'moo' lasts a certain amount of time. A short 'moo' might last time 1; a longer one might last time 24 or even 1,000,000,000 or longer (cows can really moo when they want to). No 'moo' will last more than or equal to 2^63.
It should come as no surprise that the cows have a pattern to their moos. Bessie will choose an integer c (1 <= c <= 100) that is the initial length of a moo.
After Bessie moos for length c, the cows calculate times for subsequent moos. They apply two formulae to each moo time to yield even more moo times. The two formulae are:
f1(c)=a1*c/d1+b1 (integer divide, of course) and
f2(c)=a2*c/d2+b2.
They then successively use the two new times created by evaluating f1(c) and f2(c) to create even more mooing times. They keep a sorted list of all the possible mooing times (discarding duplicates).
They are allowed to moo a total of N times (1 <= N <= 4,000,000). Please determine the length of the longest moo before they must quit.
The constants in the formulae have these constraints: 1 <= d1 < a1; d1 < a1 <= 20; 0 <= b1 <= 20; 1 <= d2 < a2; d2 < a2 <= 20; 0 <= b2 <= 20.
Consider an example where c=3 and N=10. The constants are:
a1=4 b1=3 d1=3
a2=17 b2=8 d2=2
The first mooing time is 3, given by the value of c. The total list of mooing times is:
1. c=3 -> 3 6. f2(3)=17*3/2+8 -> 33
2. f1(3)=4*3/3+3 -> 7 7. f1(28)=4*28/3+3 -> 40
3. f1(7)=4*7/3+3 -> 12 8. f1(33)=4*33/3+3 -> 47
4. f1(12)=4*12/3+3 -> 19 9. f1(40)=4*40/3+3 -> 56
5. f1(19)=4*19/3+3 -> 28 10. f1(47)=4*47/3+3 -> 65
The tenth time is 65, which would be the proper answer for this set of inputs.
Enter a description:
* Line 1: Two space-separated integers: c and N
* Line 2: Three space-separated integers: a1, b1, and d1
* Line 3: Three space-separated integers: a2, b2, and d2
Output Description:
* Line 1: A single line which contains a single integer which is the length of the Nth moo
Example 1
Entry
3 10 4 3 3 17 8 2
Export
65
A similar sequence of merging two things, through law is a priority queue just fine, but this question will time out.
However, a closer look, the sequence is a single-increasing (a> c), which do well, similar to a merger do just fine.
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 #include <map> 11 #include <math.h> 12 const int INF=0x3f3f3f3f; 13 typedef long long LL; 14 const intMOD = 1E9 + . 7 ; 15 const int MAXN = . 4 * 1e6 + 10 ; 16 the using namespace STD; . 17 18 is LL ANS [MAXN]; // ANS array is an increasing queue . 19 20 is int main () 21 is { 22 is int C, n-, A1, B1, D1, A2, B2, D2; 23 is Scanf ( " % D% D% D% D% D% D% D% D " , & C, & n-, & A1, & B1, & D1, & A2, & B2, & D2 ); 24 int CNT = . 1 ; // queue counter 25 int F1 = . 1, F2 = . 1 ; // function counter 26 is ANS [CNT ++] = c; // initial element c enqueued 27 the while (CNT <= n-) 28 { 29 LL T min = (A1 * ANS [F1] / D1 + B1 , A2 * ANS [F2] / B2 + D2); // smaller value () was taken Fl () and F2 of 30 ANS [CNT ++] = T; // the small value enqueued 31 is IF (T = ANS A1 * = [F1] / D1 + B1) // If the smaller value from the F1 (), then Fl () +. 1 pointer F1 32 F1 ++ ; 33 is IF (T == A2 * ANS [F2] / D2 B2 +) // If a smaller value from F2 (), then F2 of () +. 1 pointer F2 34 is F2 ++ ; 35 }// important to understand the contents of the while loop. Fl calculated as () or F2 of () monotonically increasing data, can be generated in such a manner the entire series 36 the printf ( " % LLD " , ANS [n-]); // final output can. 37 [ return 0 ; 38 is }