hdu Number Sequence

Number Sequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 50   Accepted Submission(s) : 10

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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

解思：f(n)=(a*f(n-1)+b*f(n-2))%7.共有7种可能取值0~6.因为a和b都是固定的所以f(n-1)和f(n-2)最多有7*7种组合。故定义一个数组，可储存52位就够了，从第3个数开始循环寻找，知道出现和第一个和第二个一样的f(n),f(n-1)便结束循环。此时便得n以内几个数一循环。具体做法详代码。

代码：

#include <iostream>
#include <cstring>

using namespace std;

int f[1000001];

int main()
{
    int a,b,n;

    while(cin>>a>>b>>n)
    {
        if(a==0&&b==0&&n==0)
            break;
        f[1]=1;
        f[2]=1;
        int xh=0;
        int i;
        for(i=3;i<=1001;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)
            {
                
                break;
            }
        }
        xh=i-2;
        n=n%xh;
        if(n==0)n=xh;
        cout<<f[n]<<endl;
    }
    return 0;
}