Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.

Output

The output contain n lines, each line output the number of result you can get .

Sample Input

3 1 11 11111

Sample Output

1 2 8

        本題的規律是斐波那契數列，在這裡就不細說了。斐波那契數列的增長速度非常快，long long的資料型別也很快就會放不下，如何表示斐波那契數列是本題的關鍵。在這裡就需要用到陣列來模擬加法，分別將兩個數的每一位取出放到兩個陣列中，在以此將兩個陣列對應的數加起來。為了解決進位的問題，我們可以將數倒著存入陣列中，然後從最低位開始相加，如果遇到進位就讓陣列中前一個數+1，自己本身與10取餘。程式碼如下：

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	int num[220][1000];
	num[1][1]=1;
	num[2][1]=2;
	for(int i=3;i<=200;i++)
	{
		for(int j=1;j<1000;j++)
		{
			num[i][j]+=num[i-1][j]+num[i-2][j];    
			if(num[i][j]>=10)
			{
				num[i][j+1]++;    //進位操作
				num[i][j]%=10;    //自己本身要與10取餘
			}
		}
	}
	int n;
	cin>>n;
	while(n--)
	{
		char a[25];
		cin>>a;
		int len=strlen(a);
		int i;
		for(i=999;i>0 && num[len][i]==0;i--);
		for(;i>0;i--)
			cout<<num[len][i];
		cout<<endl;
	}
    return 0;
}