HDU 6143 Killer Names【容斥定理】【排列组合】
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 Cim in (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 ∑i−1j=1(−1)i−jCji jn （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( Cim in + ∑i−1j=1(−1)i−jCji jn )* (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
#include0]=c[i][i]=1;
for(int j=1;j1][j-1]+c[i-1][j])%mod;
}
}
}
void mi()///求幂
{
for(int i=1;i0]=1;
for(int j=1;j1]*i%mod;
}
}
ll solve()
{
ll ans=0,all;
for(int i=1;iint 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;
} ps：去问在打排位的小班长，果不其然被骂了QAQ
不过鉴于没有打死我，还是遥远的给您笔芯(づ￣3￣)づ╭❤～
但是真的感觉组合知识全都还给涛涛了o(╯□╰)o