杭电 OJ1005Number Sequence（循环节）



Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 137754    Accepted Submission(s): 33375

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

25

 

Author

CHEN, Shunbao

#include<stdio.h>#include<string.h>int f[1000000];using namespace std;int main(){    int a,b,n;    while(~scanf("%d%d%d",&a,&b,&n))    {        if(a==0&&b==0&&n==0)break;        f[1]=1;        f[2]=1;        int i;        for(i=3; i<=1000; i++)        {            f[i]=(a*f[i-1]+b*f[i-2])%7;            if(f[i]==1&&f[i-1]==1)                break;        }        n=n%(i-2);        f[0]=f[i-2];        printf("%d\n",f[n]);    }}