HDU 6143 Killer Names【容斥定理】【排列组合】

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解决一个关于为克隆士兵命名的问题，确保名字中家族名和姓氏无相同字符，使用组合数学原理计算最大克隆数量。

题目来戳呀

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 (T≤10), denoting the number of test cases.

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

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

2
3 2
2 3

Sample Output

2
18

Source

2017 Multi-University Training Contest - Team 8

题意

famliy name 和 last name长度各为n，现有m种字符供选择，规定famliy name 和 last name中不能有一样的字符。
如：aac bbb可以，但aab acc不可以。

想法

首先，不考虑重复，全部的情况是 $C_m^i$ $i^n$ $(m-i)^n$ *→**first name每个位子有i种可能的字符，last name每个位子有m-i种可能的字符；
然鹅会有重复的，举个栗子：
可能在family name能拿a和b，last name拿剩下的其他时，出现family name只出a的情况比如a a a c c c
这和之前family name只拿a没别的出重复了因为这样也可以出现a a a c c c
为了解决去重我们考虑用容斥定理
假设有j种字符可能在前面出现过，记为 $\sum_{j=1}^{i-1}(-1)^{i-j}C_i^j$ $j^n$ （还是滚去好好看看容斥定理吧+_+）
最后用去重过的family name的总数*last name能拿的情况就是总结果了。（我他瓜竟然在这里被绕住了>_<）
所以总的公式就是( $C_m^i$ $i^n$ + $\sum_{j=1}^{i-1}(-1)^{i-j}C_i^j$ $j^n$ )* $(m-i)^n$

对着公式就可以愉快地敲代码了~虽然不知道为什么我用快速幂求n次方时tle了TAT

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=2100;
ll c[maxn][maxn],p[maxn][maxn];
int t,n,m;
void Combination()///求组合数
{
     c[0][0]=1;
     for(int i=1;i<maxn;i++)
     {
          c[i][0]=c[i][i]=1;
          for(int j=1;j<i;j++)
          {
              c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
          }
     }
}
void mi()///求幂
{
    for(int i=1;i<maxn;++i)
    {
        p[i][0]=1;
        for(int j=1;j<maxn;++j)
            p[i][j]=p[i][j-1]*i%mod;
    }
}
ll solve()
{
    ll ans=0,all;
    for(int i=1;i<m;++i)
    {
        all=p[i][n];
        int tmp=0;
        for(int j=i-1;j>0;--j)
        {
            if(tmp==0)///偶数次
            {
               all=(all-c[i][j]*p[j][n]%mod+mod)%mod;///这里要+mod，因为可能为负
            }
            else///奇数次
            {
                all=(all+c[i][j]*p[j][n]%mod)%mod;
            }
            tmp=1-tmp;
        }
        ans=(ans+c[m][i]*all%mod*p[m-i][n]%mod)%mod;
    }
    return ans;
}
int main()
{

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        Combination();
        mi();
        printf("%lld\n",solve());
    }
    return 0;
}