HDU1005 数列找规律



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
 





#include<stdio.h> #include<string.h> int main() { int a,b,n; while(scanf("%d%d%d",&a,&b,&n)==3)//scanf返回赋值成功的个数 { if(!a&&!b&&!n) break; int arr[48]; arr[1]=1; arr[2]=1; for(int i=3;i<48;i++) arr[i]=((a*arr[i-1]+b*arr[i-2])%7); printf("%d ",arr[n%48]); } return 0; }
原文地址:https://www.cnblogs.com/yfz1552800131/p/5333488.html