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).
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
解题思路：这个题目的范围比较大，如果直接用递归就应该进行一个转换，不然提交不了。它输出的值总是会小于7，所以可以设置一个循环，先判断这个数除以7的余数等于多少，然后再进行递归运算。
程序代码：
#include<cstdio>
#include<iostream>
using namespace std;
int A,B,n;
int f(int j)
{
if(j==1||j==2)
j=1;
else j=(A*f(j-1)+B*f(j-2))%7;
return j;
}
int main()
{
while(scanf("%d%d%d",&A,&B,&n)==3&&A&&B&&n)
{
int t;
t=n%49;
for(int i=1;i<=t;i++)
f(i);
printf("%d ",f(t));
}
return 0;
}
#include<iostream>
using namespace std;
int A,B,n;
int f(int j)
{
if(j==1||j==2)
j=1;
else j=(A*f(j-1)+B*f(j-2))%7;
return j;
}
int main()
{
while(scanf("%d%d%d",&A,&B,&n)==3&&A&&B&&n)
{
int t;
t=n%49;
for(int i=1;i<=t;i++)
f(i);
printf("%d ",f(t));
}
return 0;
}