HDU1005Number Sequence

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 175727 Accepted Submission(s): 43428

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

Author
CHEN, Shunbao

Source
ZJCPC2004

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一开始看见题目就拿着递归来做，忽略了将递归的层数，（在网上看见前辈们的分析，感受到了要想AC没那么简单）本题关键在于分析出 f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7的范围，由于mod算法的特性，结果只有0~6七种结果，f（n-1）和f（n-2）二者的组合只有49种

#include<stdio.h>int a, b, n, f[50];int main(){    f[1] = f[2] = 1;    while (scanf("%d%d%d",&a,&b,&n)!=EOF)    {        if (a == 0 && b == 0 && n == 0)            continue;        for (int i = 3; i <49; i++)        {            f[i] = (a*f[i - 1] + b*f[i - 2]) % 7;        }        printf("%d\n", f[n % 49]);    }    return 0；}