题解：

更新：

k=1的时候要乘n

代码：

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=5e3+5;
typedef long long ll;
typedef pair<int,int> PII;
int T;
int n,m,mod;
int fac[N][N];
int dp[N][N];
int per[N];
int power(int a,int b)
{int res=1;while(b){if(b%2==1) res=res*a%mod;a=a*a%mod;b/=2;}return res;
}
void solve()
{cin>>n>>m>>mod;for(int i=0;i<=5002;i++){fac[i][i]=1;fac[i][0]=1;}per[0]=1;for(int i=1;i<=5002;i++) per[i]=(per[i-1]*2)%mod;for(int i=1;i<=5001;i++){for(int j=1;j<=i;j++){fac[i][j]=(fac[i-1][j-1]%mod+fac[i-1][j]%mod)%mod;}}int ans=0;for(int i=1;i<=n;i++){int k1=power(2,(n-i)*(m-1)%mod)%mod;int k2=power(2,i%mod);k2=power((k2+mod-1)%mod,(m-1)%mod);ans=(ans+fac[n][i]*k1%mod*k2%mod)%mod;}int ansA=ans%mod;//A题答案int ansB2=power(2,(n-1)*(m-1)%mod)%mod*n%mod;dp[0][0]=1;for(int i=1;i<=n;i++){for(int j=1;j<=m-1;j++){dp[i][j]=i*(dp[i][j-1]+dp[i-1][j-1])%mod;}}int ansB1=0;//cout<<ansA<<endl;int cur1=1;for(int k=n;k>=2;k--){int cur=1;int kk=(per[k]-k-1+10*mod)%mod;for(int t=m-1;t>=k;t--){int res=((fac[n][k]*cur1%mod)*(fac[m-1][t]*cur%mod))%mod*dp[k][t]%mod;cur=(cur*kk)%mod;ansB1=(ansB1+res)%mod;//cout<<res<<endl;}cur1=(cur1*per[m-1])%mod;}int ansB=(ansB1+ansB2)%mod;cout<<(ansA-ansB+mod)%mod<<endl;
}
signed main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);T=1;//cin>>T;while(T--){solve();}return 0;
}