传送门：
http://acm.hdu.edu.cn/showproblem.php?pid=1005
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 198316 Accepted Submission(s): 49744
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
Sample Output
Author
CHEN, Shunbao
Source
题目意思：
输入a，b，n
根据a，b确定表达式，根据表达式确定f（n）
f（n）=(a*f(n-1)+b*f(n-2))%7
分析：
n太大了，不能递归，也不能打表
只能找规律
确定循环周期就好办了
记住循环周期随着输入是变化的
code：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,n;
int f[10000];
f[1]=1;
f[2]=1;
while(~scanf("%d %d %d",&a,&b,&n))
{
if(a==0&&b==0&&n==0)
break;
int i;
for(i=3;i<10000;i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i]==f[1]&&f[i-1]==1)//两个连续的1出现，表面循环到了一个周期的结束
break;
}
n=n%(i-2);//周期为i-2
f[0]=f[i-2];//周期结束的那个
printf("%d\n",f[n]);
}
return 0;
}
HDU1005 Number Sequence（找规律，周期是变化的）
原文：https://www.cnblogs.com/yinbiao/p/9295863.html
