題目：

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

參考https://blog.csdn.net/hurmishine/article/details/51333005

為什么循環結是49？
很簡單，因為
f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
=A * f(n - 1)mod 7 + B * f(n - 2)mod 7
7的余數可能有：0 1 2 3 4 5 6
所以A * f(n - 1)mod 7有7種情況
B * f(n - 2)mod 7有7種情況
那么整體就有7*7種情況

所以以后遇見這種mod小的，用循環結做就ok

參考代碼：

#include <iostream> using namespace std; int arr[50]; int main() {int n,a,b;arr[1]=arr[2]=1;while(cin>>a>>b>>n){if(a==0&&b==0&&n==0)break;int minn=n<50?n:50;//一個小小的優化for(int i=3; i<=minn; i++){arr[i]=(a*arr[i-1]+b*arr[i-2])%7;}cout<<arr[n%49]<<endl;}return 0; }

總結

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