博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
hdu 6143(精妙的递推)
阅读量:1916 次
发布时间:2019-04-26

本文共 3626 字,大约阅读时间需要 12 分钟。

Problem Description
> Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as the sole offspring of two Jedi Knights—Mallie and Kento Marek—who deserted the Jedi Order during the Clone Wars. Following the death of his mother, the young Marek's father was killed in battle by Darth Vader. Though only a child, Marek possessed an exceptionally strong connection to the Force that the Dark Lord of the Sith sought to exploit.
>
> When Marek died in 2 BBY, shortly after the formation of the Alliance, Vader endeavored to recreate his disciple by utilizing the cloning technologies of the planet Kamino. The accelerated cloning process—an enhanced version of the Kaminoan method which allowed for a rapid growth rate within its subjects—was initially imperfect and many clones were too unstable to take Marek's place as the Dark Lord's new apprentice. After months of failure, one particular clone impressed Vader enough for him to hope that this version might become the first success. But as with the others, he inherited Marek's power and skills at the cost of receiving his emotions as well, a side effect of memory flashes used in the training process.
>
> — Wookieepedia
Darth Vader is finally able to stably clone the most powerful soilder in the galaxy: the Starkiller. It is the time of the final strike to destroy the Jedi remnants hidden in every corner of the galaxy.
However, as the clone army is growing, giving them names becomes a trouble. A clone of Starkiller will be given a two-word name, a first name and a last name. Both the first name and the last name have exactly 
n  characters, while each character is chosen from an alphabet of size 
m . It appears that there are 
m2n  possible names to be used.
Though the clone process succeeded, the moods of Starkiller clones seem not quite stable. Once an unsatisfactory name is given, a clone will become unstable and will try to fight against his own master. A name is safe if and only if no character appears in both the first name and the last name.
Since no two clones can share a name, Darth Vader would like to know the maximum number of clones he is able to create.
 

Input
The First line of the input contains an integer 
T  (
T10 ), denoting the number of test cases. 
Each test case contains two integers 
n  and 
m  (
1n,m2000 ).
 

Output
For each test case, output one line containing the maximum number of clones Vader can create.
Output the answer 
 mod 109+7
 

Sample Input
23 22 3
 

Sample Output
2 18
 

Source
题意:
有m个字符,由你来取名字,姓和名。一个字符只能出现在姓或者名,或者不出现。姓和名的长度为n。求可以取多少个不重复的名字。
分析:其实就是涂色问题了,dp[i][j]表示长度为i用了j种颜色的方案数。所以:dp[i][j]=(dp[i-1][j]*j+dp[i-1][j-1]*(m-j+1);
代码:
#include
using namespace std;typedef long long LL;const int maxn=2005;const LL mod=1e9+7;LL dp[maxn][maxn];//长度为i用了j种颜色LL poww(int n,int m){
LL ans=1,base=n; while(m>0) {
if(m&1) ans=(ans*base)%mod; base=(base*base)%mod; m>>=1; } return ans%mod;}int main(){
int T; scanf("%d",&T); while(T--) {
int n,m; scanf("%d%d",&n,&m); dp[1][1]=m; for(int i=2;i<=n;i++) {
for(int j=1;j<=min(m,i);j++) {
dp[i][j]=((dp[i-1][j]*j)%mod+(dp[i-1][j-1]*(m-j+1))%mod)%mod; } } LL ans=0; for(int i=1;i<=m-1;i++) {
ans+=(dp[n][i]*poww(m-i,n))%mod;//左右两边方案数相乘即可。 ans%=mod; } printf("%lld\n",ans); } return 0;}

转载地址:http://iincf.baihongyu.com/

你可能感兴趣的文章
软件架构师书籍
查看>>
Java程序员到架构师的推荐阅读书籍
查看>>
LFS、BLFS、ALFS、HLFS的区别
查看>>
国外知名网站评出对程序员最具影响力的图书(附下载)
查看>>
敏捷开发与极限编程
查看>>
如何获取system()函数的pid
查看>>
iconv 文件编码转换
查看>>
QLineEdit设置ip输入规则
查看>>
Linux串口编程
查看>>
交互设计专业书籍推荐(内有部分书籍电子版下载)
查看>>
strcasestr函数
查看>>
h264 ES流文件通过计算first_mb_in_slice区分帧边界
查看>>
设置系统时间
查看>>
C++模板学习和C++ 模板套模板
查看>>
合 JSONP 和 jQuery 快速构建强大的 mashup
查看>>
自制基于地图的 mashup
查看>>
成为优秀程序员的十个有效方法
查看>>
Oracle计算时间差函数
查看>>
Linux开机启动十步骤
查看>>
source insight 字体设置
查看>>