题解 | 小苯的计算式

//暴力，纯纯暴力，直接0，c遍历一遍就完了
void solve() {
    int n, c; cin >> n >> c;
    int csize = to_string(c).size();
    int absize = n - 2 - csize;
    ll ncount = 0;
    for (int i = 0; i <= c; i++) {
        int j = c - i;
        int abnewsize = to_string(i).size() + to_string(j).size();
        if (abnewsize == absize)ncount++;
    }
    cout << ncount;