HDU_1021_费布拉切变形

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50863    Accepted Submission(s): 24079

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

 

一来想把数列所有值计算出来，发现数值增长实在太快，到几百的规模时longlong就溢出了。

然后自己推了一下发现，（a+b）%c==(a%c+b%c)%c，这应该是取模运算的性质。 然后就过了。

 

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long

int f[1000005];
int main()
{
    f[0]=1;
    f[1]=2;
    int t=2;
    while(t<1000005)
    {
        f[t]=(f[t-1]+f[t-2])%3;
        t++;
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(f[n]==0)
            printf("yes
");
        else
            printf("no
");
    }
    return 0;
}