码迷,mamicode.com
首页 > 其他好文 > 详细

HDU 1005 Number Sequence

时间:2016-02-04 06:41:04      阅读:202      评论:0      收藏:0      [点我收藏+]

标签:

Number Sequence

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

Total Submission(s): 141781    Accepted Submission(s): 34438

 

 

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

 

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int f[10000];
 6 int main()
 7 {
 8     int a,b;
 9     long long n;
10     while(cin>>a>>b>>n&&(a||b||n)){
11         int i;
12         f[1]=1,f[2]=1;
13         for(i=3;i<10000;i++){
14             f[i]=(a*f[i-1]+b*f[i-2])%7;
15             //如果有两个连着 =1,则后面的全部和前面相同,即出现了周期  
16             //这时就没必要再进行下去了,跳出循环, i-2为周期
17             if(f[i]==1&&f[i-1]==1) break;
18         }
19         n=n%(i-2);
20         f[0]=f[i-2];
21         cout<<f[n]<<endl;
22     }
23     return 0;
24 }

 

HDU 1005 Number Sequence

标签:

原文地址:http://www.cnblogs.com/shenyw/p/5180869.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!