标签：sam 数论 clu sed his process put break using
HDU 1005 Number Sequence(数论)
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
Sample Output
2
5
解题思路：因为本题的n非常大，所以通过循环直接求解是不可行的，之后我可能会想到存在某种规律，刚开始以为是每6个就循环一次，很傻，其实是在最开始的时候f(1)和f(2)都为1，那么计算到f(n)时，如果f(n)和f(n-1)的值都为1，又会和最开始一样计算下去，一直循环，n-2个为一个周期。
代码：
#include<iostream>
using namespace std;
const int INF = 10000;
int ans[INF];
int main() {
int A, B, n;
while(cin >> A >> B >> n) {
if(!A && !B && !n) break;
ans[1] = ans[2] = 1;
ans[3] = (A + B) % 7;
int t;
for(int i = 4; i < INF; i++) {
ans[i] = (A * ans[i-1] + B * ans[i-2]) % 7;
if(ans[i-1] == 1 && ans[i] == 1) {
t = i-2;
break;
}
}
int index = n%t ? n%t : t;
cout << ans[index] << endl;
}
return 0;
}
HDU 1005 Number Sequence(数论)
标签：sam 数论 clu sed his process put break using
原文地址：https://www.cnblogs.com/kindleheart/p/9634214.html
