Ipad,IPhone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
In ACM_DIY, there is one master called “Lost”. As we know he is a “-2Dai”, which means he has a lot of money.
Well, Lost use Ipad and IPhone to reward the ones who solve the following problem.
In this problem, we define F( n ) as :
Then Lost denote a function G(a,b,n,p) as
Here a, b, n, p are all positive integer!
If you could tell Lost the value of G(a,b,n,p) , then you will get one Ipad and one IPhone!
Input
The first line is one integer T indicates the number of the test cases. (T <= 100)
Then for every case, only one line containing 4 positive integers a, b, n and p.
(1 ≤a, b, n, p≤2*10
9
, p is an odd prime number and a,b < p.)
Output
Output one line,the value of the G(a,b,n,p) .
Sample Input
4
2 3 1 10007
2 3 2 10007
2 3 3 10007
2 3 4 10007
Sample Output
40
392
3880
9941
Author
AekdyCoin
Source
ACM-DIY Group Contest 2011 Spring
Recommend
notonlysuccess
先膜拜下AC
的
1004
……
这题错了好多次……orz
,没有注意到负数的取余,表示遗憾。
首先应该知道这么个公式
感谢zsasuke提出的错误,后来查了一下,这个公式的使用是有前提条件的:“c>phi(p)”,所以需要加个判断,如果c<=phi(p),说明c次方在O(logc)的范围内可以得出结果,因此可以直接暴力求解,大于的话才使用这个公式,对于这个公式的正确性我这几天试着证明下,是在别人的解题报告里看的这个公式,做FOJ1759的时候记下来的,并没有证明……
还有p是质数(题目给定的,没发现,感谢bstw111提出)所以phi(p)=p-1。
所以Fib(n)
次方就可以处理了。
对于后面那坨……
注意对于zn
的矩阵可能系数为负的,不能直接取模……
代码如下,表示写的很丑- -|||