hduoj1005,Number Sequence,水题，第一反应暴力求解，会超时，其实有捷径

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

分析：

很容易想到暴力求解，但会超时，注意分析发现如果在找到一个周期就能很容易解决问题。

用一个for循环找到最小周期。

code:

#include<iostream>#include<cstdio>#include<cmath>#define MAX 1000using namespace std;int f[MAX+5];int main(){    int a,b,n;    f[1]=1;    f[2]=1;    while(scanf("%d %d %d",&a,&b,&n),a||b||n)    {        int i,circle;        for(i=3;i<100;i++)        {            f[i]=(a*f[i-1]+b*f[i-2])%7;            if(f[i]==1&&f[i-1]==1) break;        }        circle=i-2;        printf("%d\n",f[n%circle?n%circle:circle]);    }    return 0;}