Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36549 Accepted Submission(s): 17648
Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
Output
Print the word "yes" if 3 divide evenly into F(n).
Print the word "no" if not.
Sample Input
0 1 2 3 4 5
Sample Output
no no yes no no no/* 应用求模公式 (1) (a + b) % p = (a % p + b % p) % p (2) (a - b) % p = (a % p - b % p) % p (3) (a * b) % p = (a % p * b % p) % p (4) a ^ b % p = ((a % p)^b) % p 先打表防止超时 */#include<stdio.h> int a[1000000]={7,11,0}; int main() { int i,n; for(i=2;i<1000000;i++) { a[i]=(a[i-1]%3+a[i-2]%3)%3; //(a+b)%3=(a%3+b%3)%3 } while(scanf("%d",&n)!=EOF) { if(a[n]%3==0) printf("yes\n"); else printf("no\n"); } return 0; }