DHOJ1005 HDU1005 Number Sequence

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).

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 3
1 2 10
0 0 0
  

Sample Output

  

   2
5
  

#include <iostream> using namespace std; #define M 52 int main() { int f[10000]; int a,b,n; while(cin>>a>>b>>n&&(a||b||n) ) { f[1]=1; f[2]=1; a=a%7; b=b%7; for(int i=3;i<=M;i++) { f[i]=(a*f[i-1]+b*f[i-2]) % 7; if(f[i-1]==f[3]&&f[i]==f[4]&&i>4) break; } int t=i-4; if(n<4) cout<<f[n]<<endl; else cout<<f[(n-4)%t+4]<<endl; } return 0; } /* 分析： 1，题中(1 <= A, B <= 1000, 1 <= n <= 100,000,000)，可知用蛮力肯定行不通。 2， (A * f(n - 1) + B * f(n - 2)) mod 7 =(A%7*f(n-1)+B%7*f(n-2))%7 3，因f(i)和f(i+1) 只有49种组合，因为(f(i),(i+1)均只有7种选择,就是只能是0，1，2，3，4，5，6中的一个。故周期<=49。 解题： 1，先求周期，顺便把第一个周期的f(n)求出来。 2，利用周期，直接求其余f(n)。 */