1sting

///题解：找规律，大菲波数

Problem 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

 1 #include <stdio.h>
 2 #include<iostream>
 3 #include<cstring>
 4 #define N 1001
 5 using namespace std;
 6 char a[N][N];
 7 int main()
 8 {
 9     memset(a,'0',sizeof(a)); 
10     a[1][0]='1';
11     a[2][0]='2';
12     int i,j,d=1;
13     for(i=3;i<N;i++)//i控制行数 
14     {
15         d++;   //控制列数 
16         int c=0,s;  //c为进位指数的初值 
17         for(j=0;j<=d;j++)      //j控制列数的循环
18         {
19             s=a[i-1][j]-'0'+a[i-2][j]-'0'+c;
20             c=s/10; //满十进位
21             a[i][j]=s%10+'0';   //满十的话，将舍位
22         }
23 
24     }
25     int t;
26     char s[1000];
27     scanf("%d",&t);
28     while(t--)
29     {
30         scanf("%s",s);
31         int n=strlen(s);
32         int k=N-1;
33         while(k--)
34         {
35             if(a[n][k]!='0') break;    //反向搜索第一个不为‘0’的字符
36         }
37         for(i=k;i>=0;i--)  //从后往前输出
38             printf("%c",a[n][i]);
39        printf("
");
40     }
41     return 0;
42 }

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