題目描述
對於任何正整數x,其約數的個數記作g(x)。例如g(1)=1、g(6)=4。
如果某個正整數x滿足:g(x)>g(i) 0<i<x,則稱x爲反質數。例如,整數1,2,4,6等都是反質數。
現在給定一個數N,你能求出不超過N的最大的反質數麼?
輸入輸出格式
輸入格式:
一個數N(1<=N<=2,000,000,000)。
輸出格式:
不超過N的最大的反質數。
輸入輸出樣例
輸入樣例#1:
1000
輸出樣例#1:
840
法一:打表
(騙分過樣例,打表出生省一)
#include <iostream>
#include <cstdio>
using namespace std;
int a[5000]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,554400,665280,720720,1081080,1441440,2162160,2882880,3603600,4324320,6486480,7207200,8648640,10810800,14414400,17297280,21621600,32432400,36756720,43243200,61261200,73513440,110270160,122522400,147026880,183783600,245044800,294053760,367567200,551350800,698377680,735134400,1102701600,1396755360,2001000000};
int n;
int main()
{
cin>>n;
for(int i=0;i<5000;i++)
{
if(a[i]>n) {
cout<<a[i-1]<<endl;
return 0;
}
}
}
法二:搜索+剪枝
#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;
int prime[30]={-1,2,3,5,7,11,13,17,19,23,29,31,33,37,41,43,47,53};
long long ans,maxn;
void dfs (long long x,long long y,long long z,long long n){
if (x>maxn||(x==maxn&&ans>y)) {
maxn=x;
ans=y;
}
for(int i=1;prime[i]!=0&&y*prime[z]<=n;i++){
dfs(x*i+x,y*prime[z],z+1,n);
y*=prime[z];
}
}
int main(){
long long n;
cin>>n;
ans=2147483647;
dfs(1,1,1,n);
cout<<ans;
}