HDU 1005 Number Sequence

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原文链接：http://www.cnblogs.com/zrq495/archive/2012/02/25/2368091.html

本文介绍了一种数列的定义方式，并提供了一个求解数列第n项的算法，涉及模运算的应用。

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

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int i, a, b, f[200]={0};
 5     __int64 n;
 6     while(scanf("%d%d%I64d", &a, &b, &n) && (a||b||n))
 7     {    
 8         if (n == 1 || n == 2)
 9         {
10             printf("1\n");
11             continue;
12         }
13         f[1]=1;
14         f[2]=1;
15         for (i=3; i<=200; i++)
16         {
17             f[i]=(f[i-1]*a+f[i-2]*b)%7;
18             if ((i != 3) && (f[i] == 1) && (f[i-1] == 1))
19             {
20                 f[0]=f[i-2];
21                 break;
22             }
23         }
24         printf("%d\n", f[n%(i-2)]);
25     }
26     return 0;
27 }

转载于:https://www.cnblogs.com/zrq495/archive/2012/02/25/2368091.html