Codeforces Round #452 (Div. 2) D

（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<bits/stdc++.h>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;}