HDU 1005 Number Sequence(循环周期是关键)

Number Sequence

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

Total Submission(s): 147965 Accepted Submission(s): 35965



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

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1005

思路:取模的题目结果一般都有规律.此题会误以为周期就是7,但是,这题两个数相加后在对7取模,所以就有7*7=49种情况,即,周期为:49;

AC代码:

#include <iostream>
using namespace std;
int arr[50];
int main()
{
int n,a,b;
arr[1]=arr[2]=1;
while(cin>>a>>b>>n)
{
if(a==0&&b==0&&n==0)
break;
int minn=n<50?n:50;//一个小小的优化
for(int i=3; i<=minn; i++)
{
arr[i]=(a*arr[i-1]+b*arr[i-2])%7;
}
cout<<arr[n%49]<<endl;

}
return 0;
}