BZOJ 4517: [Sdoi2016]排列计数 [容斥原理]

4517: [Sdoi2016]排列计数

题意：多组询问，n的全排列中恰好m个不是错排的有多少个

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容斥原理强行推♂倒她
$恰好m个不是错排 $

[ = ge m个不是错排 - ge m+1个不是错排inom{m+1}{m} - ge m+2个不是错排inom{m+2}{m}... \ = sum_{i=m}^n inom{n}{i} (n-i)!inom{i}{m} \ = frac{n!}{m!} sum_{i=m}^n (-1)^{i-m} frac{1}{(i-m)!} ]

预处理阶乘逆元前缀和就可以(O(1))回答了
其实错排公式也是这么推倒来的

PS:发现题解全都是用的错排公式，~~等出一道你们不知道公式的题你们再用啊~~

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N=1e6+5, P=1e9+7;
inline int read() {
	char c=getchar(); int x=0, f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}

int n, m;
ll inv[N], fac[N], facInv[N], s[N];
int main() {
	freopen("permutation.in","r",stdin);
	freopen("permutation.out","w",stdout);
	inv[1]=1; fac[0]=facInv[0]=1;
	s[0]=1;
	for(int i=1; i<N; i++) {
		if(i!=1) inv[i] = (P-P/i)*inv[P%i]%P;
		fac[i] = fac[i-1]*i%P;
		facInv[i] = facInv[i-1]*inv[i]%P;
		s[i] = (s[i-1] + ((i&1) ? -facInv[i] : facInv[i]))%P;
	}
	int T=read();
	while(T--) {
		n=read(); m=read();  
		ll ans = fac[n]*facInv[m]%P * s[n-m]%P; 
		if(ans<0) ans+=P;
		printf("%lld
", ans);
	}
}