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CodeForces - 27E

时间:2019-08-03 18:07:28      阅读:69      评论:0      收藏:0      [点我收藏+]

https://vjudge.net/problem/CodeForces-27E

求因子个数为n的最小的数
dfs枚举质因子的幂

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

int main(){
    in(n);
    ans=inf;
    dfs(0,1,1,64);
    o(ans);
    return 0;
}

 

CodeForces - 27E

原文:https://www.cnblogs.com/war1111/p/11295531.html

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