UVa 294 Divisors

Posted 2020-08-29 大四开始ACM

方法：数论 暴力

一个正整数n的因子个数d(n) 在number theory 是一个multiplicative function，有公式。利用素数筛选先求出 sqrt(1e9)内的素数，然后对范围内每一个数求解d(n), 去最大的即可。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline ‘\n‘

#define test if(1)if(0)cerr
using namespace std;
  using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

vector<ll> primes;
const int maxn = 31622; //sqrt(1e9)
bitset<maxn+1> vis(0);
void init()
{
    for (ll i = 2; i <= maxn; ++i)
    {
        if (!vis[i]) primes.pb(i);
        for (auto p : primes)
        {
            if (p * i > maxn) break;
            vis[p*i] = true;
            if (i % p == 0) break;
        }
    }
}
int u, v;
ll d(ll n)
{
    ll ret = 1;
    for (auto p : primes)
    {
        if (p > n) break;
        if (n % p) continue;
        ll cnt = 1;
        while (n % p == 0)
        {
            n /= p;
            ++cnt;
        }
        ret *= cnt;
    }
    if (n != 1) ret *= 2;
    return ret;
}
void solve()
{
    ll ans = -1, cnt = 0;
    for (ll i = u; i <= v; ++i)
    {
        ll ret = d(i);
        if (ret > cnt)
        {
            ans = i;
            cnt = ret;
        }
    }
    cout << "Between " << u << " and " << v << ", " << ans << " has a maximum of " << cnt << " divisors.\n";
}
int main()
{
    init();
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> u >> v;
        solve();
    }
}

由于范围比较小（1e4), 据说不预处理素数，对每个数直接暴力求因子也可以通过。