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
Source
这个题可以用暴力打表，但是你不一定得到正确结果，然后啊，就挖出了一组破数据，7 7 n，这个第三项就是0了，然后都是0啊。就一直爆炸，RE、TLE各种姿势都不行的
所以循环的正确开始应该是f[3]、f[4]
#include <stdio.h> int main(){ int a,b,n,i; int f[55]={0,1,1}; while(scanf("%d%d%d",&a,&b,&n),a||b||n){ f[3]=(a+b)%7; f[4]=(a*f[3]+b)%7; for(i=5;i<=n;i++){ f[i]=(a*f[i-1]+b*f[i-2])%7; if(f[i]==f[4]&&f[i-1]==f[3]){ n=(n-3)%(i-4)+3; break; } } printf("%d\n",f[n]); } return 0; }
原文：http://www.cnblogs.com/BobHuang/p/7113350.html