Number Sequence
页面数据来自(this page from)： http://acm.hdu.edu.cn/showproblem.php?pid=1005
- Time Limit: 2000/1000 MS (Java/Others)
- Memory Limit: 65536/32768 K (Java/Others)
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
Statistic | Submit | Discuss | Note
Source Code
#include <iostream>
using namespace std;
//原理： 每17个数形成一个周期 , 把下面的注释删掉，你打印出来就可以看到这个周期了
int main(){
int A,B,n;
while(cin >> A >> B >> n){
if(A==0&&B==0&&n==0)return 0;
int f[18];
f[0] = 0;
f[1]=f[2]=1;
for(int i=3;i<17;i++){
f[i] = (A * f[i - 1] + B * f[i - 2]) % 7;
//cout << "f["<< i <<"] : " << f[i]<<endl;
}
cout << f[n%17] << endl;
}
return 0;
}
时间复杂度 O(14)
周期：
0 1 1 2 3 4 1 6 0 6 6 5 4 2 6 1
这 16 个数就是一个周期
