YT14-HDU-求斐波那契数列

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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 31 2 100 0 0

Sample Output

代码如下：

#include <iostream>using namespace std;int num[10000];int main(){    int A,B,n;    num[1] = num[2] = 1;    while(cin>>A>>B>>n, A !=0|| B!=0 || n!=0)    {        int i;        for(i=3; i<10000 ; i++)        {            num[i] = (A*num[i-1] + B*num[i-2]) % 7;            if(num[i] == 1 &&num[i-1] == 1)                break;        }        n = n % (i-2);        num[0] = num[i-2];        cout << num[n] << endl;    }    return 0;}

运行结果：

代码是借鉴网上的，自己写的全都是RUNTIME ERROR。心塞，连斐波那契都不会了。

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