hdu1005 Number Sequence（斐波拉切）

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56345    Accepted Submission(s): 12756

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

 

分析：这个题目和斐波拉切很像，因为斐波拉切模上一个数有周期，因此这个题可以从周期下手；

心得：以前我做斐波拉切时计算他的周期是 if(f[i]==f[1]&&f[i-1]==f[0]) T=i-1;使用这种方法必须满足纯周期（也就是说从第一项开始）；

     可是这里周期是从第三项开始。

     所以我刚开始一直runtime error。

#include<iostream>
using namespace std;
int f[100];
int main()
{
    int a,b,n,i,t;
    while(cin>>a>>b>>n&&(a||b||n))
    {
        t=n;
        f[0]=(a+b)%7;f[1]=(a*f[0]+b)%7;//从第三项开始
        for(i=2;i<n;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i-1]==f[0]&&f[i]==f[1])
            {
                t=i-1;break;
            }
        }
        if(n<=2) cout<<1<<endl;
        else cout<<f[(n-3)%t]<<endl;
    }
    return 0;
}