BZOJ2790 [Poi2012]Distance

一个数x可以分解为∏p[i]^a[i]，设g[x]为把x分解后的sigma a[i]，则x与y的距离为g[x]+g[y]-2*g[gcd(x,y)]

用f[i]表示他给你的数中，是i的倍数的数的g的最小值，f1[i]表示次小值，n根号max(a[i])算出f值之后对每个a[i]根号枚举约数算答案即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<queue>
#include<stack>
using namespace std;
#define MAXN 100010
#define MAXM 1010
#define MAXP 1000010
#define INF 1000000000
#define MOD 1000000007
#define eps 1e-8
#define ll long long
int f[MAXP],f1[MAXP];
int a[MAXN];
int n;
bool np[MAXP];
int p[MAXP],g[MAXP],tot;
int ansv[MAXN],ans[MAXN];
void su(){
	int i,j;
	for(i=2;i<MAXP;i++){
		if(!np[i]){
			p[++tot]=i;
			g[i]=1;
		}
		for(j=1;j<=tot&&(ll)p[j]*i<MAXP;j++){
			np[p[j]*i]=1;
			g[p[j]*i]=g[i]+1;
			if(!(i%p[j])){
				break;
			}
		}
	}
}
int main(){
	int i,j;
	su();
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&a[i]);
	} 
	memset(ansv,0x3f,sizeof(ansv));
	g[0]=MAXP;
	for(i=1;i<=n;i++){
		for(j=1;j*j<=a[i];j++){
			if(!(a[i]%j)){
				if(g[a[i]]<g[a[f[j]]]){
					f1[j]=f[j];
					f[j]=i;
				}else if(g[a[i]]<g[a[f1[j]]]){
					f1[j]=i;
				}
				if(j*j!=a[i]){
					if(g[a[i]]<g[a[f[a[i]/j]]]){
						f1[a[i]/j]=f[a[i]/j];
						f[a[i]/j]=i;
					}else if(g[a[i]]<g[a[f1[a[i]/j]]]){
						f1[a[i]/j]=i;
					}
				}
			}
		}
	}
	for(i=1;i<=n;i++){
		for(j=1;j*j<=a[i];j++){
			if(!(a[i]%j)){
				int t=f[j];
				if(f[j]==i){
					t=f1[j];
				}
				int tv=g[a[i]]+g[a[t]]-2*g[j];
				if(tv<ansv[i]){
					ansv[i]=tv;
					ans[i]=t;
				}else if(tv==ansv[i]&&t<ans[i]){
					ans[i]=t;
				}
				if(j*j!=a[i]){
					t=f[a[i]/j];
					if(f[a[i]/j]==i){
						t=f1[a[i]/j];
					}
					tv=g[a[i]]+g[a[t]]-2*g[a[i]/j];
					if(tv<ansv[i]){
						ansv[i]=tv;
						ans[i]=t;
					}else if(tv==ansv[i]&&t<ans[i]){
						ans[i]=t;
					}
				}
			}
		}
	}
	for(i=1;i<=n;i++){
		printf("%d\n",ans[i]);
	}
	return 0;
}

/*
3
10 2 9 


*/