SDUT 2044 Number Sequence（循环）

Number Sequence

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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).

输入

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.

输出

For each test case, print the value of f(n) on a single line.

示例输入

1 1 3
1 2 10
0 0 0

示例输出

2
5

提示

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来源

 

示例程序

 

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;


int k[100000001];

int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF)
    {
        int x=100000001;
        if(a==0 && b==0 && n==0)
        {
            break;
        }
        k[1]=1;
        k[2]=1;
        for(int i=3; i<=n; i++)
        {
            k[i]=(a*k[i-1]+b*k[i-2])%7;
            if(k[i-1]==1 && k[i]==1)
            {
                x=i-2;
                break;
            }
        }
        printf("%d\n",k[(n)%x]);
    }
    return 0;
}