Not Fibonacci（矩阵连乘）

Not Fibonacci（矩阵连乘）

Maybe ACMers of HIT are always fond of fibonacci numbers, because it is so beautiful. Don't you think so? At the same time, fishcanfly always likes to change and this time he thinks about the following series of numbers which you can guess is derived from the definition of fibonacci number. 

The definition of fibonacci number is: 

f(0) = 0, f(1) = 1, and for n>=2, f(n) = f(n - 1) + f(n - 2) 

We define the new series of numbers as below: 

f(0) = a, f(1) = b, and for n>=2, f(n) = p*f(n - 1) + q*f(n - 2),where p and q are integers. 

Just like the last time, we are interested in the sum of this series from the s-th element to the e-th element, that is, to calculate S(n)=f(s)+f(s+1)+...+f(e);

The first line of the input file contains a single integer t (1 <= t <= 30), the number of test cases, followed by the input data for each test case. 

Each test case contains 6 integers a,b,p,q,s,e as concerned above. We know that -1000 <= a,b <= 1000,-10 <= p,q <= 10 and 0 <= s <= e <= 2147483647.

One line for each test case, containing a single interger denoting S MOD (10^7) in the range [0,10^7) and the leading zeros should not be printed.

2
0 1 1 -1 0 3
0 1 1 1 2 3

2
3

Hint: You should not use int/long when it comes to an integer bigger than 2147483647.

AC代码：

#include #include #include #include #include #include #include #include #include #include #include #include

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