Codeforces Beta Round #27 E. Number With The Given Amount Of Divisors 含n个约数最小数

http://codeforces.com/problemset/problem/27/E

RT，求含n个约数的最小的数

我们设答案p = 2^t1 * 3^t2 * …… * p^tk（其中p是第k大的质数），则必有：t1 >= t2 >= t3 >= … >= tk >= 0。
反证法证明：若不然可将{ti}由大到小排序，设形成的新有序序列是{ti’}，t1' >= t2' >= t3' >= … >= tk'；令p’ = 2^t1' * 3^t2' * …… * p^tk'，则：p' < p，但p'的约数个数却不少于p，矛盾。所以必有：t1 >= t2 >= t3 >= … >= tk。

然后就是dfs可搞，传递三个参数：构造的t序列长度，构造出的数的约数的个数，构造出的数的大小。

ULL可做

#pragma comment(linker, "/STACK:36777216")
#pragma GCC optimize ("O2")
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 1005,maxm = 1e4 + 5;
int n,pr[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
ULL ans = ~0ULL;
void dfs(int dep,int cnt,ULL res)
{
if(cnt > n)
return;
if(cnt == n){
ans = min(ans,res);
return ;
}
for(int i = 1;i <= 63;++i){
if(res > ans/pr[dep])
break;
dfs(dep+1,cnt*(i+1),res *= pr[dep]);
}
}
int main(){
RD(n);
dfs(0,1,1);
printf("%I64d\n",ans);
return 0;
}

LL可做

#pragma comment(linker, "/STACK:36777216")
#pragma GCC optimize ("O2")
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 1005,maxm = 1e4 + 5;
int n,pr[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
LL ans = 1e18;
void dfs(int dep,int cnt,LL res)
{
    if(cnt > n)
        return;
    if(cnt == n){
        ans = min(ans,res);
        return ;
    }
    for(int i = 1;i <= 63;++i){
        if(res > ans/pr[dep])
            break;
        dfs(dep+1,cnt*(i+1),res *= pr[dep]);
    }
}
int main(){
    RD(n);
    dfs(0,1,1);
    printf("%I64d\n",ans);
    return 0;
}