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 31 2 100 0 0

示例输出

25

提示

知识扩展：本类算法在数字医疗、移动证券、手机彩票、益智类解谜类游戏软件中会经常采用。

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;}