hdu 1021 Fibonacci Again(同余模定理+斐波那契数列)
2015-04-15 18:11
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Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42136 Accepted Submission(s): 20151
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
Author
Leojay
题目分析:
递推的时候模3,然后判断时只有f[i]==0的时候才能被整除
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #define MAX 1000007 using namespace std; int f[MAX]; int n; int main ( ) { f[0] = 7%3 , f[1] = 11%3; for ( int i = 2 ; i <= 1000000 ; i++ ) f[i] = (f[i-1] + f[i-2] )%3; while ( ~scanf ( "%d" , &n ) ) if ( f ) puts ( "no" ); else puts ( "yes" ); }
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