hdu 1021 Fibonacci Again ——同余的简单性质

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26755    Accepted Submission(s): 12950

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

 

先预处理。根据同余的性质求解。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
int f[1000000];
void pre(){
  f[0] = 7 % 3; f[1] = 11 % 3;
  for (int i = 2; i <= 1000000; ++i){
    f[i] = (f[i-1] % 3 + f[i-2] % 3) % 3;
  }
}
int main(void){
  int n;
#ifndef ONLINE_JUDGE
  freopen("hdu1021.in", "r", stdin);
#endif
  pre();
  while (~scanf("%d", &n)){
    if (f[n]){
      printf("no\n");
    }
    else printf("yes\n");
  }
  return 0;
}

水题……