杭电1005-Number Sequence

Number Sequence

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

Total Submission(s): 130991    Accepted Submission(s): 31832

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

 

这个题最后输出的数要针对7取余，必定是有规律的，找出循环节即可！

AC代码：

#include<cstdio>int f[1000];int main(){long long n;int a,b,i;while(scanf("%d%d%lld",&a,&b,&n),a||b||n){f[0]=1;f[1]=1;for(i=2;i<200;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;if(f[i]==1&&f[i-1]==1)break; }int x=(n-1)%(i-1);printf("%d\n",f[x]);  }  return 0; }