杭电1005 Number Sequence问题

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 116819    Accepted Submission(s): 28424

    这题简直不能叫编程题，看到网上很多代码认为循环周期为48，或是循环里必然包括1、1，这都不科学！

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

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来源： <http://acm.hdu.edu.cn/showproblem.php?pid=1005>

#include <stdio.h>
 
/**程序首先记录1到100之间的值，根据数论得到循环周期小于49（原因不详），
 * 从60之前找到循环的位置
 * 注意：循环不一定包括1、1！654 987 6163  测试得循环里不包含1、1
 
 */
int main()
{
    int f[100];
    int A,B,N;
    int begin;
    int cycle;//循环周期
    int result;//最后结果的位置
    while(1)
    {
        scanf("%d%d%d",&A,&B,&N);
        if(A==0 && B==0 && N==0)
            break;
        f[0] = 1;
        f[1] = 1;
 
        int i;
        for(i=2; i<100; i++)
        {
            f[i] = (A*f[i-1] + B*f[i-2]) % 7;
 
        }
        for(i = 3; i<100; i++)
        {
            if(f[i]==f[60] && f[i+1]==f[61])
                break;
        }
 
        cycle = 60-i;
        if(N>100)
            result = (N-1-i)%cycle+i;
        else
            result = N-1;
 
        printf("%d\n",f[result]);
 
    }
 
    return 0;
}