Number Sequence 重在找规律，48一循环



Number Sequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 27   Accepted Submission(s) : 0

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

#include<stdio.h>#include<string.h>int main(){ int mod(int a,int b,int n); int A,B,n; while(scanf("%d%d%d",&A,&B,&n)!=EOF) {  if(A==0&&B==0&&n==0)  break;  int c[100],k;  A=A%7;  B=B%7;  n=n%48;  k=mod(A,B,n);  printf("%d\n",k); } return 0;} int mod(int a,int b,int n){ if(n==1) return 1; else     if(n==2)    return 1;    else    return (a*mod(a,b,n-1)+b*mod(a,b,n-2))%7;}