题解 | 小红的连续段

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int N = 2e3+10;
typedef long long ll;
ll fac[N];
ll inv[N];

ll qpow(ll a,ll b,ll p)
{
    if(b<0)return 0;
    ll res=  1;
    for(;b;b>>=1)
    {
        if(b&1)
        {
            res = res*a%mod;
        }
        a = a*a%mod;
    }
    return res;
}

ll C(ll n,ll m)
{
    if(n<m)return 0;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    fac[0]=inv[0]=1;
    for(int i=1;i<=1001;i++)
    {
        fac[i]=fac[i-1]*i%mod;
        inv[i] = qpow(fac[i],mod-2,mod);
    }
    ll x,y;
    cin>>x>>y;
    
    if(x!=0&&y!=0)
    {
        cout<<0<<'\n';
    }
    else cout<<1<<'\n';
    for(int i=2;i<=x+y;i++)
    {

        ll res = 0;
        if(i%2==0)
        {
            ll cur = C(2,1)%mod*C(x-1,i/2-1)%mod*C(y-1,i/2-1)%mod;
            res = (res+cur)%mod;
        }
        else 
        {
            ll cur = C(x-1,i/2-1)%mod*C(y-1,i/2)%mod;
            res =(res+cur)%mod;
            cur = C(x-1,i/2)%mod*C(y-1,i/2-1)%mod;
            res = (res+cur)%mod;
        }
        

        cout<<res<<'\n';


    }

    return 0;
}

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