HDU 1005 Number Sequence

Posted 2020-08-09 刘收获

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 154923    Accepted Submission(s): 37854

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

 

 

Author

CHEN, Shunbao

 

 

Source

ZJCPC2004

 

 

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/*-----------------------f[i-1]%7与f[i-2]%7都有0,1,2,3,4,5,6七种可能的取值，所以f[i]的取值最大有49种，也就是在49次计算

内，至少出现一次循环，当f[i-1]与f[i]等于1（f[1],f[2]等于1）的时候，即为循环出现，周期cycle = i - 2；----------------------*/

 

 

 

#include<iostream>
using namespace std;

int main()
{
int cycle,i; //周期
int f[50];
int a,b,n;
f[1] = f[2] = 1;
while(cin>>a>>b>>n,a+b+n)
{
for( i = 3;i <= 49;i++)
{
f[i] = (a*f[i-1] + b*f[i-2])%7;
if(f[i] == f[i-1] && f[i] == 1) break;
}
cycle = i - 2;
n = n%cycle;
if(n==0)
cout<<f[cycle]<<endl;
else
cout<<f[n]<<endl;
}
return 0;
}