HDOJ1005：Number Sequence——递归问题

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 3 1 2 10 0 0 0 Sample Output 2 5 递归问题。首先要求出周期T48.然后他就华丽丽的变成了一道数学题。嗯……

#include
int main(){
    long a,b,n,i,func[100];
    while(scanf(“%ld%ld%ld”,&a,&b,&n) && a!=0){
        func[1]=func[2]=1;
        for(i=3;i<=48;i++)
            func[i]=(a*func[i-1]+b*func[i-2])%7;
        printf("%ldn",func[(n>48)?n%48:n]);
    }
    return 0;
}