题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=4487

题解

枚举至多有ii行jj列kk种颜色的情况，容斥一下
ans=∑i=1n∑j=1m∑k=0c(−1)n+m+c−i−j−k(k+1)i×j(ni)(mj)(ck) ans=\sum_{i=1}^n \sum_{j=1}^m \sum_{k=0}^c (-1)^{n+m+c-i-j-k}(k+1)^{i\times j} \binom{n}{i}\binom{m}{j}\binom{c}{k}

代码

#include <cstdio>

int read()
{
  int x=0,f=1;
  char ch=getchar();
  while((ch<'0')||(ch>'9'))
    {
      if(ch=='-')
        {
          f=-f;
        }
      ch=getchar();
    }
  while((ch>='0')&&(ch<='9'))
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x*f;
}

const int maxn=400;
const int mod=1000000007;

int quickpow(int a,int b,int m)
{
  int res=1;
  while(b)
    {
      if(b&1)
        {
          res=1ll*res*a%m;
        }
      a=1ll*a*a%m;
      b>>=1;
    }
  return res;
}

int n,m,c,ans,C[maxn+10][maxn+10],pow[maxn+10][maxn*maxn+10];

int main()
{
  n=read();
  m=read();
  c=read();
  C[0][0]=1;
  for(int i=1; i<=maxn; ++i)
    {
      for(int j=0; j<=i; ++j)
        {
          C[i][j]=C[i-1][j];
          if(j>0)
            {
              C[i][j]+=C[i-1][j-1];
              if(C[i][j]>=mod)
                {
                  C[i][j]-=mod;
                }
            }
        }
    }
  for(int i=1; i<=c+1; ++i)
    {
      pow[i][0]=1;
      for(int j=1; j<=n*m; ++j)
        {
          pow[i][j]=1ll*pow[i][j-1]*i%mod;
        }
    }
  for(int i=1; i<=n; ++i)
    {
      for(int j=1; j<=m; ++j)
        {
          for(int k=0; k<=c; ++k)
            {
              ans=(ans+1ll*(((n+m+c-i-j-k)&1)?(mod-1):1)*pow[k+1][i*j]%mod*C[n][i]%mod*C[m][j]%mod*C[c][k])%mod;
            }
        }
    }
  printf("%d\n",ans);
  return 0;
}