Tiling POJ 2506
2015-08-25 19:25
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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
Sample Output
Source
The UofA Local 2000.10.14
一看题,我靠,这么长,递推+高精度运算,可惜当时不会,看了题解才明白了。
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
Source
The UofA Local 2000.10.14
一看题,我靠,这么长,递推+高精度运算,可惜当时不会,看了题解才明白了。
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<queue> using namespace std; int ans[260][300]; int main() { int n,i,j; while(scanf("%d",&n)!=EOF) { memset(ans,0,sizeof(ans)); ans[0][0]=1; ans[1][0]=1; ans[2][0]=3; if(n<=2) { printf("%d\n",ans [0]); } else { int s=0; int temp=0; for(i=3; i<=n; i++) { for(j=0; j<300; j++)//把数据存入数组,高精度运算 { s=ans[i-1][j]+ans[i-2][j]*2+temp; ans[i][j]=s%10; temp=s/10; } } int flog=0; for(i=300; i>=0; i--)//从最后面输出 { if(flog!=0||ans [i]!=0)//flog是防止结果中间出现0的情况,至于结尾,应该不会出现0。 { flog=1; printf("%d",ans [i]); } } printf("\n"); } } return 0; }
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