cf918 A. Eleven

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A. Eleven

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly n characters.

Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the i-th letter of her name should be 'O' (uppercase) if i is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to n. Fibonacci sequence is the sequence f where

• f1 = 1,
• f2 = 1,
• fn = fn - 2 + fn - 1 (n > 2).

As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.

Input

The first and only line of input contains an integer n (1 ≤ n ≤ 1000).

Output

Print Eleven's new name on the first and only line of output.

Examples

input

8

output

OOOoOooO

input

15

output

OOOoOooOooooOoo

这个是cf918的A题，很简单，就是斐波那契的数列中的数字变成大写的O，其他的变成小写的o。模拟就可以了。

接下来是我写的代码，虽然和大神写的差距很大，但抓住老鼠的就是好猫，哈哈哈哈哈，加油：

#include <iostream>
#include <math.h>
#include <string.h>
#include <algorithm>

using namespace std;
int a[1005];
int c[1005];
char b[1005];
char d[1005];

int main()
{
	int n;
	scanf("%d",&n);
	if(n == 1)
	{
		printf("O");
		return 0;
	}
	if(n == 2)
	{
		printf("OO");
		return 0;
	}
	if(n == 3)
	{
		printf("OOO");
		return 0;
	}
	if(n == 4)
	{
		printf("OOOo");
		return 0;
	}
	if(n == 5)
	{
		printf("OOOoO");
		return 0;
	}
	if(n == 6)
	{
		printf("OOOoOo");
		return 0;
	}
	int y=1,z=1,s=1;
	a[0] = 1;
	a[1] = 1;
	int sum = 0;
	for(int i=2;i<=n;i++)
	{
		a[i] = a[i-1] + a[i-2];
	}
	for(int i=1;i<=n;i++)
	{
		c[y++] = i;
	}
	for(int i=1;i<=n;i++)
	{
		b[z++] = c[i] + '0';
	}
	for(int i=1;i<n-2;i++)
	{
		sum ++;
	}
		for(int i=1;i<=n;i++)
		{
			b[i] = 'o';
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=sum;j++)
			{
				if(i == a[j])
				{
					b[i] = 'O';
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			printf("%c",b[i]);
		}
	return 0;
}

接下来附上叶大神的代码：

#include<iostream>
#include<cstring>


#define MAX 10005
using namespace std;
bool UP[MAX];
int main( )
{
	int fib[MAX] = {1,1};
	int n;
	memset(UP,false,sizeof(UP));
	UP[1] = true;
	for(int i = 2; fib[i - 1] < 1005; i++){
		fib[i] = fib[i - 1] + fib[i - 2];
		//cout << fib[i] << endl;
		UP[fib[i]] = true;
	}
	cin >> n;
	for(int i = 1; i <= n; i++)
		cout << (UP[i] ? 'O' : 'o');
	cout << endl;
}

加油！