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POJ2506——Tiling

时间:2014-06-02 00:08:55      阅读:343      评论:0      收藏:0      [点我收藏+]

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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. 
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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

题目大意:给一个2*n的棋盘,用三种方块(2*2,1*2,2*1)将其铺满,为有多少种可能性。
结题思路:显然f(n)=f(n-1)+f(n-2)*2简单递推就可以。。 但是看样例就知道longlong也存不开。用string型来进行大数相加
Code:
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 1 #include<cstdio>
 2 #include<string>
 3 #include<iostream>
 4 using namespace std;
 5 string a[300];
 6 string Add(string s1,string s2)
 7 {
 8     if (s1.length()<s2.length())
 9         swap(s1,s2);
10     int i,j;
11     for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
12     {
13         s1[i]=s1[i]+(j>=0?s2[j]-0:0);
14         if(s1[i]-0>=10)
15         {
16             s1[i]=(s1[i]-0)%10+0;
17             if(i) s1[i-1]++;
18             else s1=1+s1;
19         }
20     }
21     return s1;
22 }
23 int main()
24 {
25     int i,n;
26     a[0]="1",a[1]="1",a[2]="3";
27     for (i=3; i<=250; i++)
28         a[i]=Add(Add(a[i-1],a[i-2]),a[i-2]);
29     while (cin>>n)
30        cout<<a[n]<<endl;
31     return 0;
32 }
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POJ2506——Tiling,布布扣,bubuko.com

POJ2506——Tiling

标签:des   c   style   class   blog   code   

原文地址:http://www.cnblogs.com/Enumz/p/3763291.html

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