HDOJ 1005 Number Sequence 超详细

HDOJ 1005 Number Sequence

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

一个数字序列定义如下：f（1）= 1，f（2）= 1，f（n）=（A * f（n-1）B * f（n-2））mod 7.给定A ，B和n，计算f（n）的值。
这道题是有规律的因为是取余7也就是说是0~6七个数字再加上A,B两种数据那就是7*7=49一次轮回；

#include<stdio.h>

int f(int, int, int);

int main(void)
{
    int na, nb, nn;
    while (scanf("%d %d %d", &na, &nb, &nn) && na != 0 && nb != 0 && nn != 0)
        printf("%d\n", f(na, nb, nn));
    return 0;
}

int f(int A, int B, int n)
{
    int i = 1;
    int one = 1, two = 1;
    n %= 49;//49一次轮回
    while (i <= n)
    {
        if (i == 1)
            one = 1;
        else if (i == 2)
            two = 1;
        else if (i % 2 == 1)        
            one = (A * two + B * one) % 7;
        else        
            two = (A * one + B * two) % 7;
        i++;
    }
    if (n % 2 == 1)
      return one;
    else            
      return two;
}