ACdream 1732

input

样例个数T           <=10000

每个样例一个n(2<=n<=10^8)

output

lcm(1,2,...,n)%2^32

Sample Input

5
10
5
200
15
20

Sample Output

2520
60
2300527488
360360
232792560

做法：质因分解，每个不大于n的质数不大于n的最高次幂相乘结果即为lcm(1,2,...,n)的结果

 #include <bits/stdc++.h>
 #define MAX 10000
 #define INF 100000000
 #define LL long long
 #define U unsigned
 using namespace std;
 int cas=,T;
 U int prime[],pn,*r,rn,*rr,n,*rrp;
 struct node
 {
     U int x,y;
     bool operator<(const node&a)const
     {
         return x>a.x;
     }
 };
 /*void init()        //TLE
 {
     pn=1;
     bool *vis=new bool[INF+10];
     priority_queue<node>q;
     for(int i=2;i<=MAX;i++)
         if(!vis[i])
         {
             for(int j=i*i;j<=INF;j+=i) vis[j]=1;
             q.push(node{i,i});
         }
     prime[0]=1;
     for(U int i=MAX+1,x=1;i<=INF;i++)
     {
         if(!vis[i]) prime[pn++]=i;
     }
     prime[pn++]=INF+10;
     delete []vis;
     r=new U int[pn+10];
     r[0]=1;
     for(int i=1;i<pn;i++) r[i]=r[i-1]*prime[i];
     U int res=1;
     rr=new U int[3000];
     rrp=new U int[3000];
     rr[0]=rrp[0]=1;
     rn=1;
     while(!q.empty())
     {
         node u=q.top();q.pop();
         rrp[rn]=u.x;
         rr[rn++]=res*u.y;
         if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y});
     }
     rrp[rn++]=INF+10;
     for(int i=1;i<rn;i++) rr[i]=rr[i-1]*rr[i];
 }
 */
 void init()            //最大数为10000^2，则大于10000的素数最小公倍数的最大次幂为1，小于一万时不定
 {
     bool *vis=new bool[INF+];
     pn=;
     for(U int i=;i<=INF;i++)        //O(n)筛法
     {
         if(!vis[i]) prime[pn++]=i;
         for(U int j=;j<pn&&i*prime[j]<=INF;j++)
         {
             vis[i*prime[j]]=;
             if(i%prime[j]==) break;
         }
     }
     delete []vis;            //内存不足，用完即删
     priority_queue<node>q;
     r=new U int[pn];
     for(int i=;i<pn;i++)            //prime存质数，r存第i个质数对应的结果（还没算1~10000的部分）
     {
         if(prime[i]<=MAX) { r[i]=;q.push(node{prime[i],prime[i]}); }
         else r[i]=r[i-]*prime[i];
     }
     U int res=;
     rr=new U int[];
     rrp=new U int[];
     rr[]=rrp[]=;
     rn=;
     while(!q.empty())            //rr存质数的幂次升序，rrp存底数
     {
         node u=q.top();q.pop();
         rrp[rn]=u.x;
         rr[rn++]=res*u.y;
         if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y});
     }
     rrp[rn++]=INF+;
     for(int i=;i<rn;i++) rr[i]=rr[i-]*rr[i];        //将rrp乘起来作为1~10000的结果
 }
 int main()
 {
     //printf("%d\n",sizeof(vis)+sizeof(prime));
     init();
 //    for(int i=0;i<rn;i++) printf("%u %u\n",rrp[i],rr[i]);
     //freopen("1.in","w",stdout);
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
     scanf("%d",&T);
     while(T--)
     {
         scanf("%u",&n);
         int i1=upper_bound(prime,prime+pn,n)--prime;        //查找>10000的结果
         int i2=upper_bound(rrp,rrp+rn,n)--rrp;                //查找1~10000的结果
         printf("%u\n",r[i1] * rr[i2]);
     }
     delete []r;
     delete []rrp;
     delete []rr;
     return ;
 }