poj2506——Tiling（递推+大数加）

Description

In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?
Here is a sample tiling of a 2x17 rectangle.

Input

Input is a sequence of lines,each line containing an integer number 0 <= n <= 250.
Output

For each line of input,output one integer number in a separate line giving the number of possible tilings of a 2xn rectangle.
Sample Input

2
8
12
100
200
Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251

给2x1和2x2的地板，求铺成2xn的面积有多少种排列方式
网上找的，都是大神，真的。说什么a[n]=a[n-1]+2*a[n-2]，反正我是没看出来

#include<cstdio>
#include<string>
#include<iostream>
using namespace std;
string a[300];
string Add(string s1,string s2)
{
    if (s1.length()<s2.length())
        swap(s1,s2);
    int i,j;
    for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    {
        s1[i]=s1[i]+(j>=0?s2[j]-'0':0);
        if(s1[i]-'0'>=10)
        {
            s1[i]=(s1[i]-'0')%10+'0';
            if(i) s1[i-1]++;
            else s1='1'+s1;
        }
    }
    return s1;
}
int main()
{
    int i,n;
    a[0]="1",a[1]="1",a[2]="3";
    for (i=3; i<=250; i++)
        a[i]=Add(Add(a[i-1],a[i-2]),a[i-2]);
    while (cin>>n)
       cout<<a[n]<<endl;
    return 0;
}

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