hdoj 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 133924    Accepted Submission(s): 32531

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

 

题解：1、一般这种数值特别大而又给了你一个数组，由数组的前边的项求得后边的项的题都有规律 ，注意找规律

2、f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.  每个数mod7无非有7种结果 0  1  2  3  4  5  6，我们用同余定理转换f（n）式得f(n) = (（A * f(n - 1)）mod7 +（ B * f(n - 2)）mod7) mod 7.因为我们上边说了mod7只有7种结果，那么根据上式两个对7取模的总共有7*7=49种可能，所以49为一个周期，这样我们只需要求出前49项就可以了

 

#include<stdio.h>
#include<string.h>
#define MAX 1000
int f[MAX];
void fun(int a,int b)
{
	f[1]=f[2]=1;
	for(int i=3;i<55;i++)
		f[i]=(a*f[i-1]+b*f[i-2])%7;
}
int main()
{
	int a,b,c,n,m,i,j;
	while(scanf("%d%d%d",&a,&b,&n)&&a!=0&&b!=0&&n!=0)
	{
		n=n%49;
		fun(a,b);
		printf("%d
",f[n]);
	}
	return 0;
}