HDU1005Number Sequence(找规律)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 152357 Accepted Submission(s): 37087

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

Author

CHEN, Shunbao

Source

ZJCPC2004

思路：打表找规律，a,b固定，mod7后只有0，1，2，3，4，5，6这几位数字，所以最长的周期是49，，也就是说第50个会是第一个，不排除在50前就循环

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 LL n,a,b;
 5 int main(){
 6     while(cin>>a>>b>>n){
 7         if(a==b&&b==0&&n==b)
 8         return 0;
 9         else
10         {
11             int aa[100];
12             aa[1]=aa[2]=1;
13             int i;
14             for(i=3;i<=50;i++){
15                 aa[i]=((a*aa[i-1]+b*aa[i-2]))%7;
16                 if(aa[i]==1&&aa[i-1]==1)
17                 break;
18             }
19             n=n%(i-2);
20             if(n==0)
21             cout<<aa[i-2]<<endl;
22             else
23             cout<<aa[n]<<endl; 
24         }
25     }
26 }