HDU1005

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

由于mod7,所以所有的数都是0~6这7个数的一个,那么对于f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7,f[n-1]和f[n-2]的这两个数的组合一共有7*7=49可能(就是两个数字的组合,每个数有7种可能,比如0和1,6和5,4和2),那么如果f[i-1]和f[i-2]的值为m和n,继续算下去,在49次运算之内,肯定会出现f[j-1]和f[j-2]的值为m和n,那么再之后,f[j]和f[i]的值肯定相等了,f[j+1]和f[i+1]也是相等的...

一句话,这种题肯定是有循环节存在的,不确定是49也没事,比如在500内找,这范围肯定够了.

另外要注意的就是循环不一定是从起始位置开始的,可能是从中间的某个位置i开始,到j为一个循环.

/*HDU1005 Number Sequence*/#include <iostream>using namespace std;int main() {int a,b,n,t,start,i,j,tag;int f[50];while(cin>>a>>b>>n&&a!=0) {f[0]=1; f[1]=1; start=0; tag=0;for(j=2;j<50;j++) {f[j]=(a*f[j-1]+b*f[j-2])%7;for(i=j-1;i>0;i--) {if(f[j]==f[i]&&f[j-1]==f[i-1]) {t=j-i;start=i-1;tag=1;break;}if(tag) break;}}if(n-1<start) cout<<f[n-1]<<endl;else cout<<f[start+(n-1-start)%t]<<endl;}return 0;}