Number Sequence--hdu1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128332    Accepted Submission(s): 31212

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

这个题其实应该可以考虑到有循环的，看了好多博客说直接能得出来循环节为49，但是我太笨，不明白！

我还是自己找吧！一切数据都是由f[1]=1和f[2]=1演化出来的，所以当再次出现时，一定是在循环了！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[300];
int main()
{
    int i,j,a,b,n;
    f[1]=1;
    f[2]=1;
    while(scanf("%d%d%d",&a,&b,&n),a||b||n)
    {
        int t;
        for(i=3;i<200;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i-1]==1&&f[i]==1)
            break;

        }
            i-=2;
            n=n%i;
            if(n==0)//当n=0时，说明正好是循环节的倍数，把它转化为一个循环的最后一个
            n=i;
            printf("%d
",f[n]);
    }
    return 0;
 }