HDU1005 规律题

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1005

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
(1 <= A, B <= 1000, 1 <= n <= 100,000,000).给出A,B,N,计算结果。

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

思路：一看这题，感觉暴力肯定会超时。那么狠可能是规律题。因为是mod7，所以f(n)有0-6 7钟可能，f(n+1)也是7种，所以f(n+2)最多49种。所以49之后就是规律。

#include<stdio.h>using namespace std;int main(){    int a, b, n;    int s[101] = { 0 ,1,1};    while (~scanf("%d%d%d", &a, &b, &n) && (a + b + n))    {        int m = 0,i;            for (i = 3; i < 100; i++)        {            s[i] = (a*s[i - 1] + b*s[i - 2])%7;            if (s[i] == 1 && s[i - 1] == 1)                break;        }            m = i - 2;//周期            n %= m;            if(n!=0)            printf("%d\n", s[n]);            else                printf("%d\n", s[i-2]);    }    return 0;}