不容易系列之一

Posted 2021-11-02 Jozky86

不容易系列之一

题意：

n个数，求n个人错排(全部错误)的方案数

题解：

这题地推可以求，咱们这里用二项式反演来做
设$f(i)为恰好有i个人错排$，$g(i)为最多i个人错排$，显然$g(i)=i！$
$g(n)={\sum }_{i=0}^n{C}_n^i\ast f(i)$
反演下有：
$f(n)={\sum }_{i=0}^n(-1{)}^{n-i}\ast C_n^i\ast g(i)={\sum }_{i=0}^n(-1{)}^{n-i}\ast C_n^i\ast i！$

代码：

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef ONLINE_JUDGE
#else
    startTime = clock ();
    freopen("data.in", "r", stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#else
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn=50;
ll fac[maxn];
void init(int n){
	fac[0]=1;
	fac[1]=1;
	for(int i=2;i<=n;i++){
		fac[i]=fac[i-1]*i;
	}
}
ll C(ll n,ll m){
	ll ans=1;
	for(int i=m+1;i<=n;i++)ans=ans*i;
	for(int i=1;i<=n-m;i++)ans/=i;
	return ans;
}
int main()
{
    //rd_test();
	init(45);
	ll n;
	while(scanf("%lld",&n)!=EOF){
		ll ans=0;
		for(ll i=0;i<=n;i++){
			ans+=1ll*((n-i)&1?-1:1)*C(n,i)*fac[i];
		}
		cout<<ans<<endl;
	}
    //Time_test();
}