题目链接
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
问F(n)能否被3整除.
寻找规律是件有意思的事情,发现规律是很令人鸡冻的.
n < 1,000,000, F(n)最大时已经不能用long long 表示.
能被3整除的数如123,各位数字相加也能被3整除.所以F(i)可化简为个位数字.
又F(i)的范围为[1, 9], 所以周期肯定在9*9=81之内, 又因为不会有两个9同时出现的情况,所以周期就是80.
计算前80项,然后n%80即可.
1
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9
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| #include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdlib>
#define MAX 84
#define TEST
using namespace std;
int a[MAX] = {7, 2};
void init()
{
for(int i=2; i<MAX; i++)
{
a[i] = a[i-1] + a[i-2];
if(a[i]>9)
{
a[i] = a[i]/10+a[i]%10;
}
}
}
int main()
{
#ifdef TEST
freopen("in.txt", "r", stdin);
#endif
int n,tc;
init();
while(~scanf("%d", &n))
{
if(a[n%80]%3)
{
printf("no\n");
}
else
{
printf("yes\n");
}
}
#ifdef TEST
fclose(stdin);
#endif
return 0;
}
|
