hdu 1005

hdu 1005

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 221645 Accepted Submission(s): 56087

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

构造下转移矩阵就好了

g={a,b
   1,0};

AC代码:

#include<bits/stdc++.h>
using namespace std;
int n,a,b;
int g[2][2],t[2][2],s[2][2];
void mul(int a[][2],int b[][2])
{
	memset(t,0,sizeof t);
	for(int k=0;k<2;k++)
		for(int i=0;i<2;i++)
			for(int j=0;j<2;j++)
				t[k][i]=(t[k][i]+a[k][j]*b[j][i])%7;
	for(int i=0;i<2;i++)	for(int j=0;j<2;j++)	a[i][j]=t[i][j];
}
void qmi()
{
	while(n)
	{
		if(n&1)	mul(s,g);
		n>>=1;
		mul(g,g);
	}
}
int main()
{
	while(cin>>a>>b>>n,a||b||n)
	{
		g[0][0]=a;
		g[0][1]=b;
		g[1][0]=1;
		g[1][1]=0;
		if(n==1||n==2)
		{
			cout<<1<<endl;
			continue; 
		}
		n-=3;
		for(int i=0;i<2;i++)	for(int j=0;j<2;j++)	s[i][j]=g[i][j];
		qmi();
		cout<<(s[0][0]+s[0][1])%7<<endl;
	}
	return 0;
}