杭电1005.找规律就好

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 170439    Accepted Submission(s): 42050

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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PS:本人ACM小白第一次玩CSDN，现在代码功底并不好，有错误请大家多多指教。

题意:已知：f1=f2=1，当n>=3时，f(n)=A*f(n-1)+B*f(n-2)；现给出A,B,n的值，其中 (1 <= A, B <= 1000, 1 <= n <=100,000,000)，求f(n)的值。

解题关键：找到规律，一看n的取值最大可到100,000,000，就应该要想到不能直接用数组来做，必须找到它的循环周期。

代码：

#include 
#include 
#include 
using namespace std;
int main()
{

    int i,a,b,n;
    while(~scanf("%d%d%d",&a,&b,&n)&&(a+b+n))//当（a+b+n）均为0时退出
    {
        int s[101]= {0,1,1};                //每次都初始化数组
        for(i=3; i<100; i++)
        {
            s[i]=(a*s[i-1]+b*s[i-2])%7;         //计算数组s[3]之后的值
            if(s[i]==s[2]&&s[i-1]==s[1])        //本题关键，找到循环周期。
                break;                          //找到周期后直接退出循环
        }
        //printf("%d\n",i-2);
        n=n%(i-2);                          //周期为（i-2）
        if(n!=0)
            printf("%d\n",s[n]);
        else
            printf("%d\n",s[i-2]);
    }
    return 0;
}

}