51nod 1251 Fox序列的数量 (容斥)

枚举最多数字的出现次数$k$, 考虑其他数字的分配情况.

对至少$x$种数出现$ge k$次的方案容斥, 有

$sum (-1)^xinom{m-1}{x}inom{n-(x+1)k+m-2}{m-2}$.

暴力枚举$k$和$x$, 复杂度是$O(nlogn)$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+10;
int n, m;
ll fac[N], ifac[N];
ll C(int n, int m) {
	return fac[n]*ifac[n-m]%P*ifac[m]%P;
}
int main() {
	fac[0]=ifac[0]=1;
	REP(i,1,N-1) fac[i]=fac[i-1]*i%P,ifac[i]=inv(fac[i]);
	int t;
	scanf("%d", &t);
	REP(i,1,t) {
		scanf("%d%d", &n, &m);
		if (m==1) {puts("1");continue;}
		int ans = 0;
		REP(k,1,n) REP(x,0,m-1) {
			ll p = n-(ll)(x+1)*k+m-2;
			if (p<m-2) break;
			int ret = C(p,m-2)*C(m-1,x)%P;
			if (x&1) (ans-=ret)%=P;
			else (ans+=ret)%=P;
		}
		ans = (ll)ans*m%P;
		if (ans<0) ans+=P;
		printf("%d
", ans);
	}
}