hdu1005——Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 106254    Accepted Submission(s): 25842

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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打表找规律， 只要考虑A B 在1--7之间的情况就行了

#include<map>        
#include<set>        
#include<list>        
#include<stack>        
#include<queue>        
#include<vector>        
#include<cmath>        
#include<cstdio>        
#include<cstring>        
#include<iostream>        
#include<algorithm>        
        
using namespace std; 

int xunhuan[100];

int main()
{
	int a, b;
	int n;
	while(~scanf("%d%d%d", &a, &b, &n))
	{
		if(!a && !b && !n)
			break;
		if(n == 1 || n == 2)
		{
			printf("1\n");
			continue;
		}
		if(a == 7 && b == 7)
		{
			printf("0\n");
			continue;
		}
		if((a == 1 && b == 7) || (a == 7 && b == 1))
		{
			printf("1\n");
			continue;
		}
		if(a == 2 && b == 7)
		{
			int a[3] ={4, 1, 2};
			printf("%d\n", a[(n - 1) % 3]);
			continue;
		}
		if(a == 3 && b == 7)
		{
			int a[6] = {5, 1, 3, 2, 6, 4};
			printf("%d\n", a[(n - 1) % 6]);
			continue;
		}
		if(a == 4 && b == 7)
		{
			int a[2] = {2, 4};
			printf("%d\n", a[(n - 1) % 2]);
			continue;
		}
		if(a == 5 && b == 7)
		{
			int a[6] = {3, 1, 5, 4, 6, 2};
			printf("%d\n", a[(n - 1) % 6]);
			continue;
		}
		if(a == 6 && b == 7)
		{
			int a[6] = {6, 1};
			printf("%d\n", a[(n - 1) % 2]);
			continue;
		}
		xunhuan[1] = 1;
		xunhuan[2] = 1;
		int cnt = 2;
		int fn1 = (a + b) % 7;
		int fn2 = (a * fn1 + b) % 7;
		while(fn1 != 1 || fn2 != 1)
		{
			xunhuan[++cnt] = fn1;
			xunhuan[++cnt] = fn2;
			int tmp2 = fn2;
			fn1 = (a * tmp2 + b * fn1) % 7;
			fn2 = (a * fn1 + b * tmp2) % 7;
		}
		xunhuan[0] = xunhuan[cnt];
		printf("%d\n", xunhuan[n % cnt]);
	}
	return 0;
}