反素数ant
时间限制: 1 Sec 内存限制: 128 MB
提交: 48 解决: 26
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题目描述
对于任何正整数x,其约数的个数记作g(x)。例如g(1)=1、g(6)=4。如果某个正整数x满足:g(x)>g(i) 0<i<x,则称x为反质数。例如,整数1,2,4,6等都是反质数。现在给定一个数N,你能求出不超过N的最大的反质数么?
输入
一个数N(1<=N<=2,000,000,000)。
输出
不超过N的最大的反素数。
样例输入
1000
样例输出
840
题解:
较小的质因数优于大的质因数:如7*7*2=98 g(98)不如2*2*7=28 g(28)
这个题没有想到用唯一分解定理竟然可以求约数个数(菜啊。。)
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
using namespace std;
int p[15]={1,2,3,5,7,11,13,17,19,23,29,31};
ll s[100];
ll n,m,ans=1e9,Ans;
void dfs(ll x,ll y,ll z){//第x个质数,已有y个因数,值是z
if(x>12)return;
if(y>Ans||(y==Ans&&z<ans)){
Ans=y;ans=z;
}
s[x]=0;//<--注意
while(z*p[x]<=n&&s[x]+1<=s[x-1]){
s[x]++;
z*=p[x];
dfs(x+1,y*(s[x]+1),z);
}
}
int main()
{
scanf("%lld",&n);
s[0]=1e9;
dfs(1,1,1);
printf("%lld",ans);
}
还有一种方法是直接打表一共67个数
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
using namespace std;
ll biao[]={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};
ll n;
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<68;i++)if(biao[i]>n){printf("%lld\n",biao[i-1]);return 0;}
printf("%lld\n",biao[67]);
return 0;
}