杭电 1005

递归切记：

1.如果递归的次数大于100次，都很浪费时间，故应该考虑有没有什么规律；

2.递归时应该考虑循环节的情况。

对于此题，应当这么看：

由于最后的结果是模7的，故f
可能的值只有7个，0~6，由于f
是由f[n-1]和f[n-2]决定，故最大7*7=49次一个循环，只要找到两个连续的1出现，则可以确定为一次循环开始。

Number Sequence

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

Total Submission(s): 104494 Accepted Submission(s): 25339



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

#include<iostream>
using namespace std;
int main(){
int a,b;
int n,f[50];
while(scanf("%d%d%d",&a,&b,&n)&&a&&b&&n){
int i,j;
f[1]=1;
f[2]=1;
if(n<50){
for( i=3;i<=n;i++){
f[i]=(a*f[i-1]+b*f[i-2])%7;
}
cout<<f
<<endl;
}
else{
for( i=3;i<50;i++){
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i]==1&&f[i-1]==1)
break;
}
j=n%(i-2);
if(j==0)
cout<<f[i-2]<<endl;
else
cout<<f[j]<<endl;
}
}
return 0;
}