题解 | 小苯的计算式

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll N =1e6+5, mod = 1e9 + 7, inf = 5e18;

int n,c;

void solve(){
    cin>>n>>c;

    n-=2;//符号
    string s=to_string(c);
    n-=s.size();


    ll ans=0;
    for(int i=0;i<=200000;i++){//枚举a
        s=to_string(i);

        int len=n-s.size();
        if(len<=0)continue;

        int b=c-i;
        if(b<0)continue;

        s=to_string(b);

        if(len-s.size()==0)ans++;
    }
    cout<<ans;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    int t = 1;
    //cin >> t;

    while (t--){
        solve();
    }

    return 0;
}