span i++ ive pri sstream class num sizeof name

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

如果最後一個數字和前一個合並。。f(n-2)
不合並 f(n-1)
fn = fn-2 + fn-1
大數！

#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 + 9;
#define INF 0x3f3f3f3f
#define MAXN 203

int a[MAXN][MAXN];
 
char str[MAXN];
void add(int to[], int add[])
{
    if (to[0] < add[0])
        to[0] = add[0];
    for (int i = 1; i <= to[0]; i++)
        to[i] += add[i];
    for (int i = 1; i <= to[0]; i++)
    {
        to[i + 1] += to[i] / 10;
        to[i] %= 10;
    }
    if (to[to[0] + 1] > 0)
        to[0]++;
}
void Init()
{
    memset(a, 0, sizeof(0));
    a[1][0] = 1, a[1][1] = 1;
    a[2][0] = 1, a[2][1] = 2;
    for (int i = 3; i < MAXN; i++)
    {
        add(a[i], a[i - 1]);
        add(a[i], a[i - 2]);
    }
}
void print(int p)
{
    for (int i = a[p][0]; i > 0; 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);
    }
}

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