杭电 1005 Number Sequence（）

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 101952 Accepted Submission(s): 24623

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 31 2 100 0 0

Sample Output

Author

CHEN, Shunbao

Source

ZJCPC2004

思路：数据太大，简单存入数组的办法一定超时，或者没法计算。先写出几项，然后观察规律即可得到如下规律。题目中要求解的数据，会循环出现，所以找到周期即可。

作者：洛可可

代码如下：

<span style="font-size:18px;">#include<stdio.h>int main(){    int a,b,n;    while(~scanf("%d%d%d",&a,&b,&n),a+b+n)    {    int f[110],t=n,i;    f[0]=(a+b)%7;    f[1]=(a*f[0]+b)%7;    for(i=2;i<n;i++)    {    f[i]=(a*f[i-1]+b*f[i-2])%7;    if(f[i-1]==f[0]&&f[i]==f[1])    {    t=i-1;break;                                }                    }                 if(n<=2)    printf("1\n");    else    printf("%d\n",f[(n-3)%t]);                             }    return 0;    }</span>

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