HDU 1005 Number Sequence【循环节（取模）】

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 158149    Accepted Submission(s): 38735

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

  坑

AC代码：

#include<cstdio>int a[111];int main(){int N,A,B; a[1]=a[2]=1; while(~scanf("%d%d%d",&A,&B,&N),A|B|N) {if(N<3) {printf("1\n"); continue;}if(A%7==0&&B%7==0) {        printf(N>2?"0\n":"1\n"); continue; }int T;for(int i=3;i<100;++i) {a[i]=(A*a[i-1]+B*a[i-2])%7;if(a[i-1]==1&&a[i]==1) {T=i-1-1; break; } }N%=T;if(!N) printf("%d\n",a[T]);else printf("%d\n",a[N]);} return 0;}