cf1646 属于是写啥哇啥
C 大佬的代码真的好优美

#include <bits/stdc++.h>
using namespace std;
const int INF = 100;
int main(){
  vector<long long> fact;
  for (int i = 3; i <= 14; i++){
    long long P = 1;
    for (int j = 1; j <= i; j++){
      P *= j;
    }
    fact.push_back(P);
  }
  int t;
  cin >> t;
  for (int i = 0; i < t; i++){
    long long n;
    cin >> n;
    int ans = INF;
    for (int j = 0; j < (1 << 12); j++){
      long long n2 = n;
      for (int k = 0; k < 12; k++){
        if ((j >> k & 1) == 1){
          n2 -= fact[k];
        }
      }
      if (n2 >= 0){
        ans = min(ans, __builtin_popcount(j) + __builtin_popcountll(n2));
      }
    }
    cout << ans << endl;
  }
}

#include <stdio.h>

#define N   15
#define INF 0x3f3f3f3f

int min(int a, int b) { return a < b ? a : b; }

long long ff[N + 1];

void init() {
    int n;

    ff[0] = 1;
    for (n = 1; n <= N; n++) {
        ff[n] = ff[n - 1] * n;
        if (ff[n] > 1000000000000)
            break;
    }
}

int count(long long n) {
    return n == 0 ? 0 : count(n & n - 1) + 1;
}

int main() {
    int t;

    init();
    scanf("%d", &t);
    while (t--) {
        long long n, n_;
        int b, i, ans;

        scanf("%lld", &n);
        ans = INF;
        for (b = 0; b < 1 << N; b += 4) {
            n_ = 0;
            for (i = 0; i < N; i++)
                if ((b & 1 << i) != 0)
                    n_ += ff[i];
            if (n_ <= n)
                ans = min(ans, count(b) + count(n - n_));
        }
        printf("%d\n", ans);
    }
    return 0;
}

#include <bits/stdc++.h>
#define int long long
using namespace std;
inline int cal(int x)
{
    int rtn=0;
    while(x)
        rtn+=x&1,x>>=1;
    return rtn;
}
int fac[20];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    fac[1]=1;
    for(int i=2;i<=15;i++)
        fac[i]=fac[i-1]*i;
    while(T--)
    {
        int x;
        cin >> x;
        int ans=1e9;
        for(int i=0;i<65536;i++)
        {
            int s=0,cnt=0;
            for(int j=0;j<15;j++)
            {
                if(i&(1<<j))
                {
                    s+=fac[j+1];
                    ++cnt;
                }
            }
            if(s<=x)
            {
                ans=min(ans,cnt+cal(x-s));
            }
        }
        cout << ans << "\n";
    }
    return 0;
}

using namespace std;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int TT;
  cin >> TT;
  while (TT--) {
    long long N;
    cin >> N;
    vector<long long> fact({1});
    while (fact.back() <= N) {
      fact.push_back(fact.back() * int(fact.size()));
    }
    fact.pop_back();

    int size = int(fact.size());
    int ans = 100;
    for (int i = 0; i    < (1 << size); ++i) {
      long long sum = 0;
      for (int j = 0; j < size; ++j) {
        if ((i >> j) & 1) {
          sum += fact[j];
        }
      }
      ans = min(ans, __builtin_popcount(i) + __builtin_popcountll(N - sum));
    }
    cout << ans << '\n';
  }
}

B
我以为是只需要判断a[n-1]和a[0]+a[1]的大小就行了 物大地博🤯
nmd还需要加很多 动态加a[2],a[3],动态减去a[n-2] a[n-3]

A
我人麻了 rnm亏我加ll
s/(n*n)和s/pow(n,2)不一样
pow是double的数据，可能会有精度丢失

E

#include <bits/stdc++.h>
using namespace std;
int main(){
  int n, m;
  cin >> n >> m;
  vector<int> mx(n + 1, 1);
  vector<bool> ok(n + 1, true);
  for (int i = 2; i * i <= n; i++){
    long long P = i;
    while (true){
      P *= i;
      if (P > n){
        break;
      }
      ok[P] = false;
      mx[i]++;
    }
  }
  vector<bool> used(m * 20 + 1, false);
  vector<int> cnt(21, 0);
  int c = 0;
  for (int i = 1; i <= 20; i++){
    for (int j = 1; j <= m; j++){
      if (!used[i * j]){
        used[i * j] = true;
        c++;
      }
    }
    cnt[i] = c;
  }
  long long ans = 1;
  for (int i = 2; i <= n; i++){
    if (ok[i]){
      ans += cnt[mx[i]];
    }
  }
  cout << ans << endl;
}

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,s,t) for(int i=(s),_t=(t); i<=_t; ++i)
#define DOR(i,s,t) for(int i=(s),_t=(t); i>=_t; --i)
typedef double db;
typedef long long ll;
typedef pair<int,int> pii;
#define fi first
#define se second
const int Mod=1e9+7;
const int N=1e6+50;
ll f[21];
ll g[20*N];
int vis[N];
int main() {
    int n,m;
    scanf("%d%d",&n,&m);
    FOR(i,1,20) {
        f[i]=f[i-1];
        FOR(j,1,m) if(!g[i*j]) g[i*j]=1,++f[i];
    }
    ll res=1;
    FOR(i,2,n) {
        if(vis[i]) continue;
        int c=1,j=i;
        while(1ll*j*i<=n) {
            j*=i;
            ++c;
            vis[j]=1;
        }
        res+=f[c];
    }
    printf("%lld\n",res);
    return 0;
}