hdu 2606 Renovation Problem（遞推）
阿新 • • 發佈：2018-12-14
【題目】
Renovation ProblemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 753 Accepted Submission(s): 262
Problem Description The New Year’s Day is comming ! Every family want to make theiir house looks more beautiful , so does Teddy’s . To make his own house looks different from other's , he went to the shop to buy four kinds of tiles . each kind of tiles has different size : 1*1 , 2*2 , 3*3 and 4 * 4, respectively .
Input The first line contain a T. followed by T lines ,each line contain a integer N.(1<=N <=100). Output For each case, output the ans % 19890907. Sample Input 2 1 2 Sample Output 1 5 |
【思路】
遞推。
f(n)=f(n-1)+4*f(n-2)+2*f(n-3)+f(n-4)。
還有一種情況，我們用2*2的時候會產生一個交叉的情況，如下所示： 假設'.'表示磚塊，'1'表示未鋪的地板 4*3的地板情況： ..11 11.. .... 和 .... 11.. ..11 因為這兩種情況在n>=3的情況下每行都會出現，所以我們需要列舉一遍。
【程式碼】
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <map>
#include <list>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define og(i,a,b) for(int i=a;i>=b;i--)
const int inf=0x3f3f3f3f;
using namespace std;
typedef long long ll;
const int mod=19890907;
main()
{
int n,t;
int f[105];
f[0]=1,f[1]=1,f[2]=5,f[3]=13;
go(i,4,100)
{
f[i]=f[i-1];
f[i]=(f[i]+f[i-2]*4)%mod;
f[i]=(f[i]+f[i-3]*2)%mod;
f[i]=(f[i]+f[i-4])%mod;
go(k,3,i)
f[i]=(f[i]+2*f[i-k])%mod;
}
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("%d\n",f[n]);
}
}