HDOJ 1005 Number Sequence

Number Sequence

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

Total Submission(s): 141961 Accepted Submission(s): 34496



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

Author

CHEN, Shunbao

Source

ZJCPC2004

此题的数据较大，显然暴力解出来是不可能的，肯定是需要我去寻找规律，找出是否存在循环节。

首先这题是对7取余，取余数的结果只有7种，0~6。首先f(n) = (A * f(n - 1) + B * f(n - 2))%7，我们对这个式子进行一下化简，可以得到f(n) = (A%7 * f(n - 1)%7 + B%7 * f(n - 2)%7) % 7，又因为对7取余的结果只有0~6这7种，所以f(n-1),f(n-2)只有7*7=49种结果，所以在进行49次之后肯定会出现周期，找到周期后只需要求出在循环节中的位置即可。

#include <stdio.h>
int main()
{
int i,n,A,B;
int f[50],num;
while(scanf("%d%d%d",&A,&B,&n)!=EOF&&(A||B||n))
{
f[1]=1;
f[2]=1;
for(i=3;i<=49;i++)
{
f[i]=(A*f[i-1]+B*f[i-2])%7;
if(f[i]==f[i-1]&&f[i]==1) //判断是否出现周期
break;
}
num=i-2;
n=n%num;
if(n==0)
printf("%d\n",f[num]);
else
printf("%d\n",f
);
}
return 0;
}