hdu 1021 Fibonacci Again
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斐波那契的递推;
有规律,但是我打表过了;
f[0]=7;
f[1]=11;
实际对他们%3 就是 f[0]=1. f[1]=2;正规的斐波那契了;
则有n>2 f[n]=(f[n-1]+f[n-2])%3;
递推出斐波那契数列;
则问输入n,f[n]%3==0 yes . 否则 no;
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.
Print the word "no" if not.
Sample Input
012345
#include <bits/stdc++.h>using namespace std ;long long dp[10000000];int main(){dp[0]=7%3;dp[1]=11%3;for(int i = 2;i<=1000000;i++){dp[i]=(dp[i-1]+dp[i-2])%3;}int n ;while(cin>>n){if(dp[n]%3==0) printf("yes\n");else printf("no\n");}return 0 ;}
0 0
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