写几组数据就会发现规律了啊。

。但是我是竖着看的。。

。还找了半天啊、、、

只是要用高精度来写,水题啊。就当熟悉一下java了啊。

num[i] = 2*num[i-1]-num[i-2-k]。

1513. Lemon Tale

Time limit: 1.0 second

Memory limit: 64 MB

Background

For each programmer a point comes when the last contest is lost, and it is time to retire. Even Three Programmers themselves could not escape the common lot. But the Programmers also wanted to keep
a good memory about themselves. For this noble purpose they created problems and organized extremely popular programming contests from time to time. Of course, this work was not well paid, but for true programmers a glory was more important than money.
However it is only the first half of a job to think out a brilliant problem. The second one is to create a politically correct statement for it.

Problem

The matter is the statement of some problem for the upcoming contest was written by the Third Programmer, who knew nothing about political correctness. He just wrote a story about citrus plants growing.
As a result a word "lemon" was mentioned N times in the statement.
Besides, the problem is to be looked through by famous censor Alexander K. right before the contest. And it is a known fact, that lemons remind him of oranges he hates furiously. It worries the First
and the Second Programmers greatly - they know exactly, that if a word "lemon" occurs more than Ktimes successively, the problem will be immediately disqualified from the contest.
That is why the First and the Second Programmers connived secretly to login to the server at the eve of the contest and replace some "lemons" with much more politically correct "bananas" so that the
problem could not be disqualified. How many ways are there to do it?

Input

The only line contains the integer numbers N (1 ≤ N ≤ 10000) and K (0 ≤ K ≤ N).

Output

You should output the desired number of ways.

Sample

input output
5 2
24

Hint

Let us denote a word "lemon" by a letter "L" and a word "banana" by a letter "B". So in the sample the initial sequence of words "LLLLL" might be transformed into the following politically correct sequences:
"LLBLL", "LLBLB", "LLBBL", "LLBBB", "LBLLB", "LBLBL", "LBLBB", "LBBLL", "LBBLB", "LBBBL", "LBBBB", "BLLBL", "BLLBB", "BLBLL", "BLBLB", "BLBBL", "BLBBB", "BBLLB", "BBLBL", "BBLBB", "BBBLL", "BBBLB", "BBBBL" and "BBBBB".
import java.math.BigInteger;
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in);
int n, k;
BigInteger num[] = new BigInteger [100005];
while(cin.hasNext())
{
n = cin.nextInt();
k = cin.nextInt();
BigInteger ans;
BigInteger p = BigInteger.valueOf(2);
ans = p.pow(n);
if(k == 0)
{
System.out.println(1);
continue;
}
if(n == k)
{
System.out.println(ans);
continue;
}
num[0] = BigInteger.valueOf(1);
for(int i = 1; i <= k+1; i++)
{
num[i] = p.pow(i-1);
}
for(int i = k+2; i <= n; i++)
{
num[i] = num[i-1].multiply(BigInteger.valueOf(2));
num[i] = num[i].subtract(num[i-2-k]);
}
ans = BigInteger.valueOf(0);
for(int i = n; i >= (n-k); i--)
{
ans = ans.add(num[i]);
}
System.out.println(ans);
}
}
}

URAL 1513. Lemon Tale(简单的递推)的更多相关文章

  1. URAL 1513 Lemon Tale

    URAL 1513 思路: dp+高精度 状态:dp[i][j]表示长度为i末尾连续j个L的方案数 初始状态:dp[0][0]=1 状态转移:dp[i][j]=dp[i-1][j-1](0<=j ...

  2. HDU 2569(简单的递推)

    彼岸 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submissi ...

  3. 【图灵杯 F】一道简单的递推题(矩阵快速幂,乘法模板)

    Description 存在如下递推式: F(n+1)=A1*F(n)+A2*F(n-1)+-+An*F(1) F(n+2)=A1*F(n+1)+A2*F(n)+-+An*F(2) - 求第K项的值对 ...

  4. URAL 1009 K-based numbers(DP递推)

    点我看题目 题意 : K进制的N位数,不能有前导零,这N位数不能有连续的两个0在里边,问满足上述条件的数有多少个. 思路 : ch[i]代表着K进制的 i 位数,不含两个连续的0的个数. 当第 i 位 ...

  5. Flags-Ural1225简单递推

    Time limit: 1.0 second Memory limit: 64 MB On the Day of the Flag of Russia a shop-owner decided to ...

  6. UVa 825【简单dp,递推】

    UVa 825 题意:给定一个网格图(街道图),其中有一些交叉路口点不能走.问从西北角走到东南角最短走法有多少种.(好像没看到给数据范围...) 简单的递推吧,当然也就是最简单的动归了.显然最短路长度 ...

  7. UVA10943简单递推

    题意:      给你两个数字n,k,意思是用k个不大于n的数字组合(相加和)为n一共有多少种方法? 思路:       比较简单的递推题目,d[i][j]表示用了i个数字的和为j一共有多少种情况,则 ...

  8. [LeetCode] 递推思想的美妙 Best Time to Buy and Sell Stock I, II, III O(n) 解法

    题记:在求最大最小值的类似题目中,递推思想的奇妙之处,在于递推过程也就是比较求值的过程,从而做到一次遍历得到结果. LeetCode 上面的这三道题最能展现递推思想的美丽之处了. 题1 Best Ti ...

  9. 算法技巧讲解》关于对于递推形DP的前缀和优化

    这是在2016在长沙集训的第三天,一位学长讲解了“前缀和优化”这一技巧,并且他这一方法用的很6,个人觉得很有学习的必要. 这一技巧能使线性递推形DP的速度有着飞跃性的提升,从O(N2)优化到O(N)也 ...

随机推荐

  1. Swift 4.0:访问级别(访问控制)

    基础篇 注: 下文中所提及的类和类型为Class, Enum和Struct Swift中的访问级别有以下五种: open: 公开权限, 最高的权限, 可以被其他模块访问, 继承及复写. public: ...

  2. 计数排序(counting-sort)

    计数排序是一种稳定的排序算法,它不是比较排序.计数排序是有条件限制的:排序的数必须是n个0到k的数,所以计数排序不适合给字母排序.计数排序时间复杂度:O(n+k),空间复杂度:O(k),当k=n时,时 ...

  3. C语言实现简化的正则表达式

    语法: 正则表达式和待匹配字符串都是一行 "^" 标记正则表达式的开始 "$" 标记正则表达式的结束 "*" 匹配前面的子表达式零次或多次 ...

  4. Zabbix分布式配置

    Zabbix是一个分布式监控系统,它可以以一个中心点.多个分节点的模式运行,使用Proxy能大大的降低Zabbix Server的压力,Zabbix Proxy可以运行在独立的服务器上,安装Zabbi ...

  5. eclipse/myeclipse中js/java的自动提示只有4个字符怎么解决

    https://blog.csdn.net/LinBM123/article/details/80450690

  6. 小贝_php源代码安装

    PHP安装  一.本文档相关文件下载 二.php安装 一.本文档相关文件下载 1.php下载地址: http://php.net/downloads.php (备注: 本文档下载的是php版本号为ph ...

  7. 蓝的成长记——追逐DBA(18):小机上WAS集群故障,由一次更换IP引起

    原创作品.出自 "深蓝的blog" 博客,欢迎转载,转载时请务必注明出处.否则追究版权法律责任. 深蓝的blog:http://blog.csdn.net/huangyanlong ...

  8. 操作指定文件格式的10个Perl CPAN模块

    在Perl开发中,非常可能会碰到一些不同格式的文件--XML.PDF.CSV及RSS文件等,和一些不同的二进制数据格式.Perl应用程序须要操作这些文件,对它们进行读写. 此时.能够求助于全面Perl ...

  9. angularjs 标签指令

    <!DOCTYPE HTML> <html ng-app="myApp"> <head> <meta http-equiv="C ...

  10. Traversing a list

    The most common way to traverse the elements of a list is with a for loop. The syntax is the same as ...