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
题意:
给出A ,B,n,求f(n)的值
思路:
根据乘法分配律可以将 f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7 变为 f(n)=(A f(n-1)mod7)+(Bf(n-2)mod7) ,那么A * f(n-1)mod7的结果有七种可能(0,1,2,3,4,5,6),同理,B*f(n-2)mod7也是,所以最终f(n)的结果只有7 *7=49种可能,所以这就是我们所熟知数学中数特别大,但是有循环规律的简单题喽,如果想不通的话,建议输出60或100个结果进行观察,理解这个,代码就很容易实现了,思路明白代码就好办了!
代码:
#include<stdio.h>
int f[10000];
int main()
{
f[1]=1;
f[2]=1;
int a,b;
long long n,i;
while(~scanf("%d %d %lld",&a,&b,&n)&&a+b+n)
{
for(i=3; i<=100; i++)
{
f[i]=(a*(f[i-1])+b*(f[i-2]))%7;
}
//for(i=1; i<=100; i++)
printf("%d\n",f[n%49]);
}
return 0;
}