HDU 1005 Number Sequence
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我就在想到底有没有循环节,最后我决定,本题目是有循环节的,因为只要出现两个相邻的数值相同,那么他就会重置f(n)的值进入循环,至于是不是1,是不能确定的。如果出现5和5相邻,是不是就从5开始循环了 ,这 也是确定的。这题目真是有点乱。原解释矩阵快速幂。不过还搞不懂,加油。
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
#include<stdio.h>#include<iostream>using namespace std;int a[10005];int A, B, n;int main(){ a[1] = a[2] = 1; while(scanf("%d%d%d", &A, &B, &n), A || B || n) { int i; for(i=3; i<10000; i++) { a[i] = (A*a[i-1] + B*a[i-2]) % 7; if(a[i] == 1 && a[i-1] == 1) break; } n = n % (i-2); a[0] = a[i-2]; cout<<a[n]<<endl; } return 0;}
0 0
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