杭电ACM 1005 Number Sequence

Number Sequence

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

Total Submission(s): 170412    Accepted Submission(s): 42042

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  

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#include<iostream>
using namespace std;
typedef long long ll;
const int mod=7;
#define N 2
struct matrix
{
	ll a[2][2];
};
matrix multiple(matrix x, matrix y, int n)
{
	matrix tmp;
	memset(tmp.a, 0, sizeof(tmp.a));
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			for (int k = 0; k < n; k++)
				tmp.a[i][j] = (tmp.a[i][j] + (x.a[i][k] * y.a[k][j]) % mod) % mod;
	return tmp;
}
int main(void)
{
	matrix res;
	int a, b, n;
	cin >> a >> b >> n;
	while (a != 0 && b != 0 && n != 0)
	{
		memset(res.a, 0, sizeof(res.a));
		for (int i = 0; i<2; i++) {
			res.a[i][i] = 1;
		}
		if (n == 1 || n == 2)
			cout << 1 << endl;
		else
		{
			n -= N;
			matrix tmp = { a,b,
				           1,0 };
			while (n)
			{
				if (n & 1) res= multiple(res, tmp, N);
				n >>= 1;
				tmp=multiple(tmp, tmp, N);
				
			}
			cout << (res.a[0][0] + res.a[0][1]) % mod << endl;
		}
		cin >> a >> b >> n;
	}
}
           

知识来源：

http://www.cnblogs.com/CXCXCXC/p/4641812.html   http://blog.csdn.net/wust_zzwh/article/details/52058209