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
题意:有一个斐波拉且数列,求是否f[n]能整除3
思路1:由于数据过大,会超出任何类型的表达范围,所以我用了同余定理
(a+b) % c = a %c + b % c,能整除的部分已经被我们去除,我们考虑的是不能整除的那部分,从而值不会超出表示范围
#include<stdio.h>
const int maxn = 1000000+1;
int f[maxn] = {0};
int main()
{
int n,i;
int flag;
f[0] = 7;f[1] = 11;
for(i = 2;i <= maxn;++i) //可用(a+b) mod c = a mod c + b mod c
{
f[i-1] %= 3;
f[i-2] %= 3;
f[i] = f[i-1]+f[i-2];
}
while(scanf("%d",&n)!=EOF)
{
flag = 0;
if(f[n]%3 == 0)
flag = 1;
if(flag)
printf("yes\n");
else
printf("no\n");
}
return 0;
}
思路:从第二项起,每四项输出结果重复一遍
#include<iostream>
using namespace std;
int main()
{
int n;
while (cin>>n)
{
if ((n - 2) % 4 != 0)
cout << ("no") << endl;
else
cout << ("yes") << endl;
}
return 0;
}