组合数取模（lucas定理+CRT合并）（AC）

 #include<bits/stdc++.h>
 #define re register
 #define int long long
 using namespace std;
 int T,n,m,mod,ans=;
 inline int read(){
     re int a=,b=;re char ch=getchar();
     while(ch<''||ch>'')
         b=(ch=='-')?-:,ch=getchar();
     while(ch>=''&&ch<='')
         a=(a<<)+(a<<)+(ch^),ch=getchar();
     return a*b;
 }
 inline int qpow(re int a,re int b,re int p){
     re int res=;a%=p;
     for(;b;b>>=){
         if(b&)res=res*a%p;
         a=a*a%p;
     }
     return res%p;
 }
 inline int exgcd(re int a,re int b,re int &x,re int &y){
     re int d=a;
     if(!b){x=,y=;return a;}
     d=exgcd(b,a%b,y,x);
     y-=(a/b)*x;
     return d;
 }
 inline int inv(re int a,re int m){
     re int x,y;
     re int d=exgcd(a,m,x,y);
     return d==?(m+x%m)%m:-;
 }
 inline int mul(re int n,re int pi,re int pk){
     if(!n)return ;
     re int ans=;
     if(n/pk){
         for(re int i=;i<=pk;i++)
             if(i%pi) ans=ans*i%pk;
         ans=qpow(ans,n/pk,pk);
     }
     for(re int i=;i<=n%pk;i++)
         if(i%pi) ans=ans*i%pk;
     return ans*mul(n/pi,pi,pk)%pk;
 }
 inline int C(re int n,re int m,re int p,re int pi,re int pk)
 {
     if(m>n)return ;
     re int a=mul(n,pi,pk),b=mul(m,pi,pk),c=mul(n-m,pi,pk);
     re int k=,ans;
     for(re int i=n;i;i/=pi) k+=i/pi;
     for(re int i=m;i;i/=pi) k-=i/pi;
     for(re int i=n-m;i;i/=pi) k-=i/pi;
     ans=a*inv(b,pk)%pk*inv(c,pk)%pk*qpow(pi,k,pk)%pk;
     ans=ans*(p/pk)%p*inv(p/pk,pk)%p;
     return ans;
 }
 inline int lucas(re int m,re int n,re int p){
     re int x=p,ans=;
     for(re int i=;i<=p;i++){
         if(!(x%i)){
             re int pk=;
             while(!(x%i))pk*=i,x/=i;
             ans=(ans+C(n,m,p,i,pk))%p;
         }
     }
     return ans;
 }
 signed main(){
     T=read(),mod=read();
     n=read(),m=read();
     if(mod==||T<n+m){puts("");return ;}
     for(re int k=m;k<=(T+m-n)/;k++)
         ans+=lucas(k,T,mod)%mod*lucas(k-m,T-k,mod)%mod*lucas((T+m+n-(k<<))/,T-(k<<)+m,mod)%mod;
     printf("%lld\n",ans%mod);
     return ;
 }

 #include<bits/stdc++.h>
 #define re register
 #define ll long long
 using namespace std;
 ll T,n,m,mod,ans=;
 inline ll read(){
     re ll a=,b=;re char ch=getchar();
     while(ch<''||ch>'')
         b=(ch=='-')?-:,ch=getchar();
     while(ch>=''&&ch<='')
         a=(a<<)+(a<<)+(ch^),ch=getchar();
     return a*b;
 }
 inline void exgcd(re ll &x,re ll &y,re ll a,re ll b)
 {
     if(!b){x=,y=;return;}
     exgcd(x,y,b,a%b);
     ll tmp=x;x=y;
     y=tmp-a/b*y;
 }
 inline ll inv(re ll a,re ll p)
 {
     re ll x,y;
     exgcd(x,y,a,p);
     return x;
 }
 inline ll qpow(re ll a,re ll b,re ll p){
     re ll res=;a%=p;
     for(;b;b>>=){
         if(b&)res=res*a%p;
         a=a*a%p;
     }
     return res%p;
 }
 inline ll fac(re ll n,re ll p,re ll tp)
 {
     if(!n) return ;
     re ll f=,res=;
     for(re ll i=;i<tp;i++)
     {
         if(i%p) (f*=i)%=tp;
         if(i==n%tp) res=f;
     }
     f=qpow(f,n/tp,tp);
     return fac(n/p,p,tp)*f%tp*res%tp;
 }
 inline ll C(re ll m,re ll n,re ll p,re ll tp)
 {
     re ll ct=;
     for(re ll i=m;i;i/=p) ct+=i/p;
     for(re ll i=n;i;i/=p) ct-=i/p;
     for(re ll i=m-n;i;i/=p) ct-=i/p;
     return fac(m,p,tp)*inv(fac(n,p,tp),tp)%tp*inv(fac(m-n,p,tp),tp)%tp*qpow(p,ct,tp)%tp;
 }
 inline ll CRT(re ll m,re ll n,re ll p,re ll tp)
 {
     return C(m,n,p,tp)*(mod/tp)%mod*inv(mod/tp,tp)%mod;
 }
 inline ll lucas(re ll n,re ll m,re ll p)
 {
     ll ans=;
     for(re ll i=;i*i<=p;i++)
     {
         if(p%i==)
         {
             re ll tp=;
             while(p%i==) tp*=i,p/=i;
             (ans+=CRT(m,n,i,tp))%=mod;
         }
     }
     if(p!=) (ans+=CRT(m,n,p,p))%=mod;
     return (ans%mod+mod)%mod;
 }
 signed main(){
     T=read(),mod=read();
     n=read(),m=read();
     if(mod==||T<n+m){puts("");return ;}
     for(re int k=m;k<=(T+m-n)/;k++)
         ans+=lucas(k,T,mod)%mod*lucas(k-m,T-k,mod)%mod*lucas((T+m+n-(k<<))/,T-(k<<)+m,mod)%mod;
     printf("%lld\n",ans%mod);
     return ;
 }

以下是ac代码

附上新的公式推导方式（经实测可AC）

设k为向上走的步数（纯的，不包括向下），那么向下走的步数为k-m

即C（T,k）×C(T-k，k-m);

由上式可推出 向右的与向左的步数之和为T-2*k+m,而向右的比向左的多n，利用和差公式，易得，向右的为（T-2*k+m+n)/2;

且k>=m,（T-2*k+m+n)/2>=n;可得，k->[m,(T+m-n)/2];

故整体公式为

　　∑（k->[m,(T+m-n)/2]）（C（T,k）×C(T-k，k-m)×C（T-2*k+m,（T-2*k+m+n)/2）);

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 using namespace std;
 #define QAQ 100010
 #define re register
 #define int long long
 #define ll long long
 #define fup(i,a,b) for(re int i=a;i<=b;++i)
 #define fdn(i,a,b) for(re int i=a;i>=b;--i)
 int n,m,mod,t;
 ll ans;
 int fac[QAQ],inv[QAQ];
 int tot,p[QAQ],w[QAQ];
 inline int quick_fk_me(re int a,re int b,re int pk){
     int ans=;
     for(;b;b>>=){
         if(b&)ans=1ll*ans*a%pk;
         a=1ll*a*a%pk;
     }
 }
 void exgcd(re ll a,re ll b,re ll &x,re ll &y){
     if(b==){x=,y=;return;}
     exgcd(b,a%b,x,y);
     re int z=x;
     x=y,y=z-a/b*y;
 }
 inline ll crt(){
     re ll ans=,lcm=,x,y;
     fup(i,,tot)lcm*=p[i];
     fup(i,,tot){
         re ll tmp=lcm/p[i];
         exgcd(tmp,p[i],x,y);
         x=(x%p[i]+p[i])%p[i];
         ans=(ans+tmp*x*w[i])%lcm;
     }
     return (ans+lcm)%lcm;
 }
 inline void Get_fac_and_inv(re int top,re int pk){
     fac[]=fac[]=inv[]=inv[]=;
     fup(i,,top)fac[i]=1ll*fac[i-]*i%pk;
     fup(i,,top)inv[i]=(pk-pk/i)*inv[pk%i]%pk;
     fup(i,,top)inv[i]=1ll*inv[i-]*inv[i]%pk;
 }
 int CC(re int n,re int m,re int pk){
     if(m>n)return ;
     return fac[n]*inv[m]%pk*inv[n-m]%pk;
 }
 int lucas(re int n,re int m,re int pk){
     if(!m)return ;
     return CC(n%pk,m%pk,pk)*lucas(n/pk,m/pk,pk)%pk;
 }
 inline bool divide(int n){
     re int top=sqrt(n);
     fup(i,,top){
         if(n%i)continue;
         p[++tot]=i;
         while(n%i==)n/=i;
     }
     if(n>)p[++tot]=n;
     return tot==;
 }
 main(){
     scanf("%lld%lld",&t,&mod);
     scanf("%lld%lld",&n,&m);
     if(n<)n=-n;
     if(m<)m=-m;
     re int st=m,ed=(t+m-n)>>;
     if(divide(mod)){
         Get_fac_and_inv(min(t,mod),mod);//什么纱吊错 妈耶 不求阶乘？？！！
         fup(k,st,ed)
         ans=(ans+lucas(t,k,mod)*lucas(t-k,k-m,mod)%mod*lucas(t-*k+m,(t-*k+m-n)>>,mod)%mod)%mod;
         printf("%lld\n",ans);
         return ;
     }
     fup(i,,tot){
         re int pk=p[i];
         Get_fac_and_inv(min(t,pk),pk);
         fup(k,st,ed)
         w[i]=(w[i]+lucas(t,k,pk)*lucas(t-k,k-m,pk)%pk*lucas(t-*k+m,(t-*k+m-n)>>,pk)%pk)%pk;
     }
     printf("%lld\n",crt());
 }

 #include<bits/stdc++.h>
 #define re register
 #define int long long
 using namespace std;
 int T,n,m,mod,ans=,tot=,fac[],pi,pk,p;
 inline int read(){
     re int a=,b=;re char ch=getchar();
     while(ch<''||ch>'')
         b=(ch=='-')?-:,ch=getchar();
     while(ch>=''&&ch<='')
         a=(a<<)+(a<<)+(ch^),ch=getchar();
     return a*b;
 }
 inline int qpow(re int a,re int b){
     re int res=;a%=pk;
     for(;b;b>>=){
         if(b&)res=res*a%pk;
         a=a*a%pk;
     }
     return res%pk;
 }
 inline int getcnt(re int x){
     return x?x/pi+getcnt(x/pi):;
 }
 inline int getsum(re int x){
     return x?((x/pk?qpow(fac[pk],x/pk):)*fac[x%pk]%pk*getsum(x/pi)%pk):;
 }
 inline int C(re int x,re int y){
     if(y>x)return ;
     re int a0=getsum(x),b0=getcnt(x);
     re int a1=getsum(y),b1=getcnt(y);
     re int a2=getsum(x-y),b2=getcnt(x-y);
       b0-=b1+b2;
       if(b0>=tot)return ;//含质因子的个数
       return a0*qpow(a1,p-)%pk*qpow(a2,p-)%pk*qpow(pi,b0)%pk;
 }
 inline int work(){
     re int res=,a,b,c,d;
       fac[]=;
       for(re int i=;i<=pk&&i<=T;i++)
         fac[i]=(i%pi)?fac[i-]*i%pk:fac[i-];//处理阶乘
       a=,b=n,c=(T-n-m)>>,d=(T+m-n)>>;//各方向初值
       while(c>=){
         res=(res+C(T,a+b)*C(a+b,a)%pk*C(c+d,c)%pk)%pk;
         ++a,++b,--c,--d;
      }//处理组合数，上下左右的道路可能性
       return res;
 }
 signed main(){
     T=read(),mod=read(),n=read(),m=read();
     n=(n<)?-n:n;m=(m<)?-m:m;//处理负数
     re int l=mod;
     for(re int i=;l>;i++){
         if(i*i>l)i=l;//分解质因数
         if(l%i==){
             pi=i,pk=,tot=;
             while(l%pi==)
                 l/=pi,pk*=pi,++tot;
             p=pk/pi*(pi-);
             (ans+=work()%mod*(mod/pk)%mod*qpow(mod/pk,p-)%mod)%=mod;//CRT
         }
     }
     printf("%lld\n",ans);
     return ;
 }

另有 学习笔记“组合数取模”推荐