标签：equal int ike possible limit 一个 描述 prope eterm
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.
Partial feedback will be provided on the first 50 submissions. MEMORY LIMIT: 64MB
满月的时候，和狼一样，牛们也在月光下叫.他们从不嚎叫，而是哞叫.
每次哞叫都有一个时长，可能是1秒，可能是；10^9秒或更久，牛们真的非常能叫.当然，没有 哞叫时长会超过或等于2^63.
牛们的哞叫可以找到规律，这并不奇怪.贝茜会选择一个整数100)来作为初始时 长.之后，牛们根据两条公式确定更多的时长.这两条公式
f1(c)=a1*c/d1+b1整除
f2(c)=a2*c/d2+b2.
牛们用这两条公式不断地迭代、计算，算得大量的时长.然后她们将这些时长排序，剔除重 复的时长，最后取前N(l<N< 4000000)个整数为她们N次哞叫的时长.请你计算，第N次哞叫 的时长是多少.公式中的常量均为整数，满足下列关系： 1 <= d1 < a1; d1 < a1 <= 20; 0 <= b1 <= 20; 1 <= d2 < a2; d2 < a2 <= 20; 0 <= b2 <= 20.
* 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
* Line 1: A single line which contains a single integer which is the length of the Nth moo
3 10 4 3 3 17 8 2
65
#include<iostream>
#include<cstdio>
using namespace std;
int n,h,a,b,c,x,y,z,i,j,k;
long long q[4000005],tmpa,tmpb;
int main(){
scanf("%d %d",&h,&n);
scanf("%d %d %d",&a,&b,&c);
scanf("%d %d %d",&x,&y,&z);
q[1]=h;
j=k=1;
for(i=2;i<=n;i++){
tmpa=1ll*a*q[j]/c+b;
tmpb=1ll*x*q[k]/z+y;
if(tmpa==tmpb){
j++;
k++;
q[i]=tmpa;
}
else
if((tmpa>tmpb&&tmpb>0)||tmpa<0){
k++;
q[i]=tmpb;
}
else
if((tmpb>tmpa&&tmpa>0)||tmpb<0){
j++;
q[i]=tmpa;
}
}
cout<<q[n];
}
P2942 [USACO09MAR]Moon哞哞叫Moon Mooing
标签：equal int ike possible limit 一个 描述 prope eterm
原文地址：https://www.cnblogs.com/xiongchongwen/p/11781386.html
