hdoj--1005--Number Sequence（规律题）

Number Sequence

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 137047    Accepted Submission(s): 33211
[/align]

[align=left]Problem Description[/align]
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).

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, print the value of f(n) on a single line.

 

[align=left]Sample Input[/align]

1 1 3
1 2 10
0 0 0

 

[align=left]Sample Output[/align]

2
5

 

[align=left]Author[/align]
CHEN, Shunbao
 

[align=left]Source[/align]
ZJCPC2004

 

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规律题，48一循环

 #include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[48];
int main()
{
int A,B,n;
while(scanf("%d%d%d",&A,&B,&n)!=EOF)
{
if(A==0&&B==0&&n==0) break;
memset(a,0,sizeof(a));
a[2]=a[1]=a[3]=1;
int t;
for(int i=3;i<=48;i++)
a[i]=(A*a[i-1]%7+B*a[i-2]%7)%7;
printf("%d\n",a[n%48]);
}
return 0;
}