POJ2506-Tiling

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

解答
该题算法很简单，主要用到的递推式是

rst[i] = rst[i-1] + 2*rst[i-2];

需要注意两点

• 结果太大，用long long 和double都存不下，用一个string来进行加法运算。
• n = 0时， 返回1

#include<cstdio>#include<string>#include<iostream>using namespace std;string rst[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 n;    rst[0] = "1";    rst[1] = "1";    rst[2] = "3";    for(int i = 3; i < 300; i++)        //rst[i] = rst[i-1] + 2*rst[i-2];        rst[i] = Add(Add(rst[i-1], rst[i-2]), rst[i-2]);    while(scanf("%d", &n) != EOF)    {        cout << rst[n] << endl;    }    return 0;}