【hdoj1005】Number Sequence

Number Sequence

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

Total Submission(s): 158167    Accepted Submission(s): 38744


[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

 

[align=left]Recommend[/align]
JGShining
 规律是48个一个循环，直接打表前48个，再取余一下就好了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mod 7
using namespace std;
typedef long long LL;
const int N = 100;
int ans
;
void init(int A,int B){
ans[1]=1,ans[2]=1;
for(int i=3;i<=48;i++){
ans[i]=(A*ans[i-1]%mod+B*ans[i-2]%7)%7;
}
}
int main(){
LL A,B,n;
while(scanf("%lld%lld%lld",&A,&B,&n),A||B||n){
init(A,B);
printf("%d\n",ans[n%48]);
}
return 0;
}