hdu 1005 矩阵乘法
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Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37082 Accepted Submission(s): 7790
Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input
1 1 31 2 100 0 0
Sample Output
25
Author
CHEN, Shunbao
Source
ZJCPC2004
Recommend
JGShining
很久没做题,一下子还没想到矩阵乘法,用普通做法是要TLE,所以只能用矩阵乘法了。
公式是
[f1 f2] ( [1 1] )
*
[ 0 b
1 a ] ^(n-2)
后面这个部分用矩阵乘法做就可以了
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