1005：Number Sequence（hdu，数学规律题）

 

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

#include<iostream>
using namespace std;
int main()
{
    int f[4000];
    int a, b, n;
    while (cin >> a >> b >> n, a != 0 && b != 0 && n != 0)
    {
        int i;
        f[1] = 1, f[2] = 1;
        for (i = 3; i < 500; i++)
        {
            f[i] = (a*f[i - 1] + b * f[i - 2]) % 7;
            if (f[i] == 1 && f[i - 1] == 1) break;
         }
        i -= 2;
        if (i > n)
        {
            cout << f[n] << endl;
            continue;
        }
        n = n % i;
        if (n == 0) n = i;
        cout << f[n] << endl;
    }
    return 0;
}

大佬代码：http://www.cnblogs.com/kuangbin/archive/2011/07/26/2117381.html

#include<stdio.h>
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int A,B,i;
    long n;
    int f[201];
    f[1]=f[2]=1;
    while(scanf("%d %d %ld",&A,&B,&n))
    {
        if(A==0&&B==0&&n==0) break;
        int cnt=0;
        for(i=3;i<=200;i++)//打表找到周期 
        {
            f[i]=(A*f[i-1]+B*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)break;
            if(f[i]==0&&f[i-1]==0){cnt=1;break;}//这里有个小陷阱，如果A=7，B=7则后面都为0了 
        }
        if(cnt){printf("0
");continue;}
        if(i>n){printf("%d
",f[n]);continue;}
        i-=2;//i为周期
        n%=i;
        if(n==0)n=i;
        printf("%d
",f[n]);    
    } 
    return 0;   
    
}    

这题完全参考大佬的代码改良；

凉凉；；；；；；；；；

数据量小，一些不合适的没在测试数据不在里面