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hdu-1005 Number Sequence

2024-09-16 08:26:35   214人阅读

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
2
5

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;}

 

hdu-1005 Number Sequence


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