1sting（大数递归）

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

题意：两个1凑在一起可以合成2，问可以合成几种数字

比如111可以有21、12、111三种

大数的递归

代码（C）

#include<stdio.h>
#include<string.h>
int a[1010][1010];//因为是大数，一维数组并不能储存数字，所以用二维，后面用来分别储存大数的每个位数
int main()
{
int q,m,n,t;
int ans,r,s,i;
r=0;
memset(a,0,sizeof(a));
a[1][1]=1;
a[2][1]=2;
for(i=1;i<=997;i++)//大数的加法
{
for(int j=1;j<=1010;j++)
{
s=a[i][j]+a[i+1][j]+r;
a[i+2][j]=s%10;
r=s/10;
}
}
scanf("%d",&t);
while(t--)
{
char str[201];
scanf("%s",&str);
n=strlen(str);
if(n==1)  printf("1\n");
else if(n==2)  printf("2\n");
else
{

for(i=1010;i>=1;i--) if(a
[i])break; //当a
[i]为0时跳出，此时的i即答案有几位数，进入下一步的输出
for(;i>=1;i--)
printf("%d",a
[i]);
printf("\n");
}
}
return 0;
}