hdu-1005 Number Sequence

题目链接：

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

题目：

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 170367 Accepted Submission(s): 42027

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

题意概括：输入三个数A B n，根据题目给出的公式f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.计算f(n)。

解题思路：我们从公式得出一个结论：无论n有多大，f(n)总为一个不大于7的数，而且，f(n)是根据f(n-1)和f(n-2)得到的，所以f(n)必成一个循环数组，所以这个数组最坏的可能的循环节为7*7=49，并且49一定是这个数组的循环节，所以可以根据打表前49个数据，并对49取余得到相应的答案。这就是抽屉原理，详情请自行百度。

AC代码：

# include <stdio.h>

int a,b;

int f(int x,int y)
{
    int sum;
    sum=(a*x+b*y)%7;
    return sum;
}

int main ()
{
    int i,n,ret,l,m[1010];
    m[1]=1; m[2]=1;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF)
    {
        if(!a&&!b&&!n)
            break;
        for(i=3;i<100;i++)//根据抽屉原理，循环节必在前五十个中 
            m[i]=f(m[i-1],m[i-2]);
            
        printf("%d\n",m[n%49]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/love-sherry/p/6745136.html