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
1 #include <bits/stdc++.h> 2 #define MAX 10000 3 #define INF 100000000 4 #define LL long long 5 #define U unsigned 6 using namespace std; 7 int cas=1,T; 8 U int prime[6000000],pn,*r,rn,*rr,n,*rrp; 9 struct node 10 { 11 U int x,y; 12 bool operator<(const node&a)const 13 { 14 return x>a.x; 15 } 16 }; 17 /*void init() //TLE 18 { 19 pn=1; 20 bool *vis=new bool[INF+10]; 21 priority_queue<node>q; 22 for(int i=2;i<=MAX;i++) 23 if(!vis[i]) 24 { 25 for(int j=i*i;j<=INF;j+=i) vis[j]=1; 26 q.push(node{i,i}); 27 } 28 prime[0]=1; 29 for(U int i=MAX+1,x=1;i<=INF;i++) 30 { 31 if(!vis[i]) prime[pn++]=i; 32 } 33 prime[pn++]=INF+10; 34 delete []vis; 35 r=new U int[pn+10]; 36 r[0]=1; 37 for(int i=1;i<pn;i++) r[i]=r[i-1]*prime[i]; 38 U int res=1; 39 rr=new U int[3000]; 40 rrp=new U int[3000]; 41 rr[0]=rrp[0]=1; 42 rn=1; 43 while(!q.empty()) 44 { 45 node u=q.top();q.pop(); 46 rrp[rn]=u.x; 47 rr[rn++]=res*u.y; 48 if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y}); 49 } 50 rrp[rn++]=INF+10; 51 for(int i=1;i<rn;i++) rr[i]=rr[i-1]*rr[i]; 52 } 53 */ 54 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 55 { 56 bool *vis=new bool[INF+10]; 57 pn=0; 58 for(U int i=2;i<=INF;i++) //O(n)筛法 59 { 60 if(!vis[i]) prime[pn++]=i; 61 for(U int j=0;j<pn&&i*prime[j]<=INF;j++) 62 { 63 vis[i*prime[j]]=1; 64 if(i%prime[j]==0) break; 65 } 66 } 67 delete []vis; //内存不足，用完即删 68 priority_queue<node>q; 69 r=new U int[pn]; 70 for(int i=0;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） 71 { 72 if(prime[i]<=MAX) { r[i]=1;q.push(node{prime[i],prime[i]}); } 73 else r[i]=r[i-1]*prime[i]; 74 } 75 U int res=1; 76 rr=new U int[3000]; 77 rrp=new U int[3000]; 78 rr[0]=rrp[0]=1; 79 rn=1; 80 while(!q.empty()) //rr存质数的幂次升序，rrp存底数 81 { 82 node u=q.top();q.pop(); 83 rrp[rn]=u.x; 84 rr[rn++]=res*u.y; 85 if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y}); 86 } 87 rrp[rn++]=INF+10; 88 for(int i=1;i<rn;i++) rr[i]=rr[i-1]*rr[i]; //将rrp乘起来作为1~10000的结果 89 } 90 int main() 91 { 92 //printf("%d ",sizeof(vis)+sizeof(prime)); 93 init(); 94 // for(int i=0;i<rn;i++) printf("%u %u ",rrp[i],rr[i]); 95 //freopen("1.in","w",stdout); 96 //freopen("1.in","r",stdin); 97 //freopen("1.out","w",stdout); 98 //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC); 99 scanf("%d",&T); 100 while(T--) 101 { 102 scanf("%u",&n); 103 int i1=upper_bound(prime,prime+pn,n)-1-prime; //查找>10000的结果 104 int i2=upper_bound(rrp,rrp+rn,n)-1-rrp; //查找1~10000的结果 105 printf("%u ",r[i1] * rr[i2]); 106 } 107 delete []r; 108 delete []rrp; 109 delete []rr; 110 return 0; 111 }