hdu1005 Number Sequence（寻找循环节）

主题链接：

pid=1005">huangjing

题意：

就是给了一个公式，然后求出第n项是多少。。。

思路：

题目中n的范围实在是太大，所以肯定直接递推肯定会超时，所以想到的是暴力打表，找循环节，可是也不是那么easy发现啊，所以这时候分析一下，由于最后都会mod7，所以总共同拥有7X7总情况，即A 0，1,2,3,4,5,6,7。B也是如此，所以循环节为49，这么这个问题就攻克了。

。。

题目：

Number Sequence

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

Total Submission(s): 105674    Accepted Submission(s): 25691

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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代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

#define mod 7

const int maxn=1000+10;
int a[maxn];
int A,B,n;

int main()
{
    while(~scanf("%d%d%d",&A,&B,&n))
    {
        if(A==0&&B==0&&n==0)  return 0;
        a[1]=1,a[2]=1;
        for(int i=3;i<=100;i++)
             a[i]=(A*a[i-1]+B*a[i-2])%mod;
        printf("%d\n",a[n%49]);
    }
    return 0;
}