hdu 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129135    Accepted Submission(s): 31436

 

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

 

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通过打表发现循环是有规律的，但是没有发现循环节是多少，在网上查看的时候，看到了这段解释。

对于公式 f[n] = A * f[n-1] + B * f[n-2]; 后者只有7 * 7 = 49 种可能，为什么这么说，因为对于f[n-1] 或者 f[n-2] 的取值只有 0,1,2,3,4,5,6 这7个数，A，B又是固定的，所以就只有49种可能值了。由该关系式得知每一项只与前两项发生关系，所以当连续的两项在前面出现过循环节出现了，注意循环节并不一定会是开始的 1，1 。

 

题意：题意很简单，题目中已经给了一个公式，输入数据A，B，n，A,B为公式A,B的值，n指的是输出数组中第n个数字。

 

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a,b,n;
int s[1005]= {0,1,1};
void init()
{
    /*  for(int i=3; i<=100; i++)
      s[i]=(a*s[i-1]+b*s[i-2])%7;
      for(int i=1;i<=100;i++)
          cout<<s[i]<<" ";
     cout<<endl;*/        //打表发现数据有规律
    for(int i=3; i<=48; i++)
        s[i]=(a*s[i-1]+b*s[i-2])%7;    //发现规律中最大循环次数
}
int main()
{
    int i,j;
    while(~scanf("%d%d%d",&a,&b,&n))
    {
        if(a==0&&b==0&&n==0)
            break;
        init();
        printf("%d
",s[n%48]);   //对最大循环次数取余
    }
}