hdu 1005 Number Sequence【规律题】

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 138304 Accepted Submission(s): 33514

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 31 2 100 0 0

Sample Output

25

因为题目要求所有的数据都会对7 取余，那么在49次数据之内，肯定会出现循环的情况！

然后就是求循环开始的地点以及循环的长度了，不过个人感觉自己的程序有漏洞，虽然AC了......

#include<stdio.h>int a,b,n,x[105],bg,len;void fun()//找循环开始的位置和循环的周期 {x[1]=x[2]=1;for(int i=3;i<100;++i){x[i]=(a*x[i-1]+b*x[i-2])%7;for(int j=2;j<i;++j){if(x[i]==x[j]&&x[i-1]==x[j-1]){bg=j;//记录第一个位置 len=i-j;//长度 return;}}}}int main(){//freopen("shuju.txt","r",stdin);while(scanf("%d%d%d",&a,&b,&n),a|b|n){fun();if(n>bg){n=(n-bg)%len+bg;}printf("%d\n",x[n]);}return 0;}