AGC002F - Leftmost Ball

Solution

设$f_{i,j}$表示放$i$个白球，确定了$j$个颜色的球的位置的方案数。

有两种转移：

1. 放白球，$f_{i,j}->f_{i+1,j}$
2. 放完一种颜色的球，$f_{i,j}->f_{i,j+1}$，方案数为$(n-j)\ast \left(\genfrac{}{}{0ex}{}{nk-i-j\ast (k-1)-1}{k-2}\right)$。

相当于当前情况下，没选过的第一个必选以保证方案不重复。

时间复杂度$O(n^2)$。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=2005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int fac[MAXN*MAXN],inv[MAXN*MAXN],f[MAXN][MAXN];
int upd(int x,int y) { return x+y>=mods?x+y-mods:x+y; }
int C(int x,int y) { return 1ll*fac[x]*inv[y]%mods*inv[x-y]%mods; }
int quick_pow(int x,int y)
{int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;
}
void Init(int n)
{fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;inv[n]=quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mods;
}
signed main()
{int n=read(),k=read();if (k==1) { puts("1"); return 0; }Init(n*k),f[0][0]=1;for (int i=0;i<=n;i++) for (int j=0;j<=i;j++){if (i<n) f[i+1][j]=upd(f[i+1][j],f[i][j]);if (j<i) f[i][j+1]=upd(f[i][j+1],1ll*f[i][j]*(n-j)%mods*C(n*k-j*(k-1)-i-1,k-2)%mods);}
//	for (int i=0;i<=n;i++)
//		for (int j=0;j<=n;j++) cout<<i<<" "<<j<<":"<<f[i][j]<<endl;printf("%d\n",f[n][n]);return 0;
}