Educational Codeforces Round 187 (Rated for Div. 2)-CF-Edu-187-赛后总结（较大数据量下 vector 套 vector 的调和级数因数分解会出问题）（涉及到累加器的，用 for 循环，不要用 while 循环） · znzryb
## 基本情况
两道题目，哎，还好这个B做的快是吧，绷不住了。
当然了，小号其实也无所谓。嗯，其实也还好不过之后呢，我觉得还是把这个号打上去以后，还是用这个号打。其实，你一直没分数，其实也没什么意思，就是那种比较刺激的那种感觉是吧。
OK，我们说回来。这个D为什么交了5发还是没过呢？主要是因为就是这个D，它的数据量比较大。使用Victor victor或者victor的这种调和级数写法呢，它这个时间会超。可以使用vis 数组标记所有 a 的倍数。这样子可以避免卡常。
```cpp
ll cnt_un_divide=0,cnt_all_divide=0;
vector<char> vis(N+M+2);
for (auto a:A) {
for (ll i=1;i<=N+M;++i) {
if (i*a>N+M) {
break;
}
vis[i*a]=true;
}
}
```
然后为什么标题里面提了不要用 while 循环？是因为涉及累加的逻辑（在 D 的对拍暴力程序当中），直接 continue 以后，while 循环的这个累加就崩掉了。
A 就是一个向上取整除法，没什么好说的。
::: A 源代码
```cpp
/**
* Problem: A. Towers of Boxes
* Contest: Educational Codeforces Round 187 (Rated for Div. 2)
* Judge: Codeforces
* URL: https://codeforces.com/contest/2203/problem/A
* Created: 2026-02-25 22:35:14
* Author: Gospel_rock
* My blog: https://znzryb.com/
*
* Powered by AutoCp https://github.com/Pushpavel/AutoCp
*/
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define lson(o) (o<<1)
#define rson(o) (o<<1|1)
#define SZ(a) ((long long) a.size())
#define debug(var) cerr << #var <<" = ["<<var<<"]"<<"\n";
#define debug1d(a) \
cerr << #a << " = ["; \
for (int i = 0; i < (int)(a).size(); i++) \
cerr << (i ? ", " : "") << a[i]; \
cerr << "]\n";
#define debug2d(a) \
cerr << #a << " = [\n"; \
for (int i = 0; i < (int)(a).size(); i++) \
{ \
cerr << " ["; \
for (int j = 0; j < (int)(a[i]).size(); j++) \
cerr << (j ? ", " : "") << a[i][j]; \
cerr << "]\n"; \
} \
cerr << "]\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using DB = double;
// using i128 = __int128;
using CD = complex<double>;
static constexpr ll MAXN = (ll)1e6+10, INF = (1ll<<61)-1;
static constexpr ll mod = 998244353; // (ll)1e9+7;
static constexpr double eps = 1e-8;
const double pi = acos(-1.0);
ll lT,testcase;
/*
*
*/
void Solve() {
ll N,M,D;
cin >> N >> M >> D;
ll ans=(N+D/M)/(D/M+1);
cout<<ans<<"\n";
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef LOCAL
cout.setf(ios::unitbuf); // 无缓冲流，方便我们调试
#endif
cin >> lT;
for (testcase=1;testcase<=lT;++testcase)
Solve();
return 0;
}
/*
*/
```
:::
B 也简单，就是猜一下，我猜，只有这个 digit 总和小于 10 的情况下，题设条件才成立。注意前导 0，不要不小心进行操作后出现前导 0。
可以在排序前，对第一个数减 1，让其被操作后至少保留 1。
```cpp
sn[0]--;
sort(all(sn),greater<>());
```
::: B 源代码
```cpp
/**
* Problem: B. Beautiful Numbers
* Contest: Educational Codeforces Round 187 (Rated for Div. 2)
* Judge: Codeforces
* URL: https://codeforces.com/contest/2203/problem/B
* Created: 2026-02-25 22:39:19
* Author: Gospel_rock
* My blog: https://znzryb.com/
*
* Powered by AutoCp https://github.com/Pushpavel/AutoCp
*/
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define lson(o) (o<<1)
#define rson(o) (o<<1|1)
#define SZ(a) ((long long) a.size())
#define debug(var) cerr << #var <<" = ["<<var<<"]"<<"\n";
#define debug1d(a) \
cerr << #a << " = ["; \
for (int i = 0; i < (int)(a).size(); i++) \
cerr << (i ? ", " : "") << a[i]; \
cerr << "]\n";
#define debug2d(a) \
cerr << #a << " = [\n"; \
for (int i = 0; i < (int)(a).size(); i++) \
{ \
cerr << " ["; \
for (int j = 0; j < (int)(a[i]).size(); j++) \
cerr << (j ? ", " : "") << a[i][j]; \
cerr << "]\n"; \
} \
cerr << "]\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using DB = double;
// using i128 = __int128;
using CD = complex<double>;
static constexpr ll MAXN = (ll)1e6+10, INF = (1ll<<61)-1;
static constexpr ll mod = 998244353; // (ll)1e9+7;
static constexpr double eps = 1e-8;
const double pi = acos(-1.0);
ll lT,testcase;
/*
*
*/
void Solve() {
ll N;
cin >> N;
string sn=to_string(N);
ll sum=0;
for (auto ch:sn) {
sum+=ch-'0';
}
sn[0]--;
sort(all(sn),greater<>());
ll ans=0;
if (sum<10) {
cout<<ans<<"\n";
return;
}
for (auto ch:sn) {
ll num=ch-'0';
sum-=num;
++ans;
if (sum<10) {
cout<<ans<<"\n";
return;
}
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef LOCAL
cout.setf(ios::unitbuf); // 无缓冲流，方便我们调试
#endif
cin >> lT;
for (testcase=1;testcase<=lT;++testcase)
Solve();
return 0;
}
/*
*/
```
:::
## 心得感悟