ZOJ2562

ZOJ2562
https://vjudge.net/problem/11781/origin
<=n的且因子数最多的那个数
做法：同因子数取最小，dfs更新答案

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf ~0
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(unsigned long long i=a;i<=b;++i)

using namespace std;
unsigned long long n,ans,now;
unsigned long long prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
void in(unsigned long long &x){
    unsigned long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(unsigned long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

void dfs(unsigned long long depth,unsigned long long num,unsigned long long cnt,unsigned long long up){
    if(num>=now&&cnt<=ans)
        return;
    if(ans<cnt){
        ans=cnt;
        now=num;
    }
    if(ans==cnt&&now>num)
        now=num;
    For(i,1,up){
        if(num*prime[depth]>n) return;
        dfs(depth+1,num*=prime[depth],cnt*(i+1),i);
    }
}

int main(){
    while(scanf("%llu",&n)!=EOF){
        ans=0;
        now=0;
        dfs(0,1,1,64);
        o(now);
        p('\n');
    }
    return 0;
}