编程开发Shovel Sale 规律+暴力

（场上没出，此代码参考并得到了瑞神指导 ）

规律+暴力

首先判不能组成的末位9的情况；

然后判后面连续的9的个数k，然后枚举末位含有k个连续的9的数，每个数组成个数加入到答案中

#include
typedef long long ll;

using namespace std;

ll a[22] = {0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999LL, 9999999999LL};
ll b[22] = {0, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000LL, 10000000000LL};


int main() {

    ll n, sum;

    cin >> n;

    sum = n + n - 1;
    int p = 0;
    while(sum >= a[p]) ++p;
    --p;

    if(p == 0) {
        cout << (n - 1) * n / 2 << endl;
        return 0;
    }

    ll ans = 0, cur = a[p];

    for(int i = 0; i < 10; ++i) {
        sum = n + n - 1;
        if(sum < cur) break;

        if(n >= cur) {
            ans += (cur - 1) / 2;
        } else {
            ll m = n + n - cur + 1;
            if(m > 1) {
                ans += m / 2;
            }
        }
        cur += b[p];
    }
    cout << ans << endl;

    return 0;
}