杭电ACM1005 C做法

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 173466    Accepted Submission(s): 42860

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

 

脑子锈了不少，简单的题目都要想不少时间了。。。参考了其他大神的一些想法，贴个网址

http://blog.csdn.net/middle544291353/article/details/7723427

余数出现的循环周期为7*7=49次，因此我直接将需要输出的n%49，这样只要计算一次49个数，之后出现的到之前数组找就行，有看到大神用long int类型，我用的是int类型也AC了，直接贴代码吧

#include
int main()
{
	int a,b,n;
	int f[100];
	f[1] = 1,f[2] = 1;
	while(scanf("%d%d%d",&a,&b,&n)!=EOF&&(a!=0)&&(b!=0)&&(n!=0))
	{
		if(n==1||n==2) printf("1\n");
		else
		{
			n %= 49;
			for(int i=3;i<=n;i++)
			{
				f[i] = (a*f[i-1]+b*f[i-2])%7;	
			}
			printf("%d\n",f[n]);
		}
	}
	return 0;
}