题解

化简 题给公式有
an- a1 ==k

可行解则是
a1 .... an

枚举a1
可得an的范围
中间段的可能取值，显而易见可通过快速幂得出

累加a1的所有可行解即可

#include<bits/stdc++.h>

using namespace std;

#define ll long long 
#define ull unsigned long long 
#define qios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define inf INT_MAX
#define il inline

const ll mod=1e9+7;
ll qpow(ll a, ll b)
{
    ll res=1;
    while (b)
    {
        if (b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}
signed main()
{
    ll n, k;
    cin>>n>>k;
    ll ans=0;
    for (int i=1;i<=9;++i)
        if (i+k<=9 and i+k>=0)
            ans=(ans+qpow(10, n-2)%mod)%mod;
    std::cout << ans << std::endl;
    return 0;
}