1sting 大数 递推

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. 

InputThe 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. 
OutputThe 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

如果最后一个数字和前一个合并。。f(n-2)
不合并 f(n-1)
fn = fn-2 + fn-1
大数！

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<fstream>
#include<set>
#include<memory>
#include<bitset>
#include<string>
#include<functional>
using namespace std;
typedef long long LL;
const int MAXN = 1e6 + ;
#define INF 0x3f3f3f3f
#define MAXN 203

int a[MAXN][MAXN];
char str[MAXN];
void add(int to[], int add[])
{
    if (to[] < add[])
        to[] = add[];
    for (int i = ; i <= to[]; i++)
        to[i] += add[i];
    for (int i = ; i <= to[]; i++)
    {
        to[i + ] += to[i] / ;
        to[i] %= ;
    }
    if (to[to[] + ] > )
        to[]++;
}
void Init()
{
    memset(a, , sizeof());
    a[][] = , a[][] = ;
    a[][] = , a[][] = ;
    for (int i = ; i < MAXN; i++)
    {
        add(a[i], a[i - ]);
        add(a[i], a[i - ]);
    }
}
void print(int p)
{
    for (int i = a[p][]; i > ; i--)
        printf("%d", a[p][i]);
    printf("\n");
}
int main()
{
    int n, t;
    scanf("%d", &n);
    Init();
    while (n--)
    {
        scanf("%s", &str);
        t = strlen(str);
        print(t);
    }
}