hdu 1005 找规律（好坑！！）

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

25

这题评测机有问题，周期用中间变量会出错，for循环数据不当会出错

#include<iostream>#include<cstring>#include<string>#include<cmath>#include<cstdio>#include<vector>#include<stack>#include<map>#include<algorithm>#define inf 0x3f3f3f3f#define ll long longusing namespace std;int f[10010];int main(){    int a,b,n;    while(cin>>a>>b>>n)    {        if(a==0&&b==0&&n==0)            return 0;    f[1]=1;    f[2]=1;    int i;    for(i=3;i<=10000;i++)    {        f[i]=(a*f[i-1]+b*f[i-2])%7;        if(f[i]==1&&f[i-1]==1)            break;    }    n=n%(i-2);    f[0]=f[i-2];    cout<<f[n]<<endl;    }    return 0;}