Hdu6143 Killer Names（2017多校第8场）

> 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.
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> — 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.

The First line of the input contains an integer  T  ( T≤10 ), denoting the number of test cases. 

Each test case contains two integers  n  and  m  ( 1≤n,m≤2000 ).

For each test case, output one line containing the maximum number of clones Vader can create.

Output the answer   mod 109+7

  
  

   
   2
3 2
2 3
  
  

  
  

   
   2 
18
  
  

2017 Multi-University Training Contest - Team 8

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 题目的意思是用m个字母取构成两个长度为n的字符串，两个字符串不能有公共的字母，问最大可能种数

思路：先预处理出一个数组，存确定用i的字母取构成长度为n的字符串方案数，这个可以递推出来，然后取枚举前一个字符串取多少个，用组合数学搞一搞

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

LL a[2005];
const LL mod=1e9+7;
LL mypow(LL x,LL y)
{
    LL ans=1;
    while(y>=1)
    {
        if(y&1)
            ans=(ans*x)%mod;
        x=(x*x)%mod;
        y>>=1;
    }
    return ans;
}

LL c[2022][2022];
void chs()
{
    c[0][0] = 1;
    for(int i = 1; i < 2022; i++)
    {
        c[i][0] = c[i][i] = 1;
        for(int j = 1; j < i; j++)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
}

int main()
{
    int T;
    LL m,n;
    chs();
    for(scanf("%d",&T); T--;)
    {
        scanf("%lld%lld",&n,&m);
        for(int i=1; i<m; i++)
        {
            a[i]=mypow(i,n);
            for(int j=1; j<i; j++)
                a[i]=((a[i] - c[i][j]*a[j] % mod) % mod + mod) % mod;
        }
        LL ans = 0;
        for(int i=1; i<m; i++)
        {
            for(int j = 1; j <= m - i; j++)
                ans = (ans + c[m][i] * a[i] % mod * c[m - i][j] % mod * a[j] % mod) % mod;
        }
        printf("%lld\n", (ans % mod + mod) % mod);
    }
    return 0;
}