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


[align=left]Problem Description[/align]
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

[align=left]Input[/align]
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

[align=left]Output[/align]
Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

[align=left]Sample Input[/align]

0
1
2
3
4
5

[align=left]Sample Output[/align]

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
==0)
printf("yes\n");
else
printf("no\n");
}
return 0;
}