LightOJ 1289 LCM from 1 to n

1289 - LCM from 1 to n

Given an integer n, you have to find

lcm(1, 2, 3, ..., n)

lcm means least common multiple. For example lcm(2, 5, 4) = 20, lcm(3, 9) = 9, lcm(6, 8, 12) = 24.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (2 ≤ n ≤ 10⁸).

Output

For each case, print the case number and lcm(1, 2, 3, ..., n). As the result can be very big, print the result modulo 2³².

Sample Input	Output for Sample Input
5 10 5 200 15 20	Case 1: 2520 Case 2: 60 Case 3: 2300527488 Case 4: 360360 Case 5: 232792560

Solution:

首先,lcm(1,2,...,n)为2^k1*3^k2*...*px^kx,pi为小于等于n的质数,pi^ki<=n且pi^(ki+1)>n

所以,先预处理出1e8范围内的质数,及其前缀积.

设pro[i]为∏p[j](1<=j<=i)

ans=1;
for(i=1;i<=31;++i)
{
　　找到p[j]i<=n且p[j+1]i>n;
　　ans*=pro[j];
}

Code:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<bitset>
using namespace std;
#define uint unsigned int
#define ull unsigned long long
const uint MAXN=1e8;
const uint MAXTOT=5761455;
const uint MAXPOW=31;
const uint INF=0xffffffff;
uint t,CASE;
bitset<MAXN+10> vis;
uint p[MAXTOT+10],tot;
uint pro[MAXTOT+10];
void GetP()
{
	for(uint i=2;i<=MAXN;++i)
	{
		if(!vis[i])p[++tot]=i;
		for(uint j=1;j<=tot&&(ull)p[j]*i<=MAXN;++j)
		{
			vis[p[j]*i]=true;
			if(i%p[j]==0)break;
		}
	}
}
uint QuickPow(uint x,uint y)
{
	ull res=1;
	while(y--)
	{
		res*=x;
		if(res>INF)return INF;
	}
	return res;
}
void init()
{
	GetP();
	pro[0]=1;
	for(uint i=1;i<=tot;++i)pro[i]=pro[i-1]*p[i];
}
uint n;
int main()
{
	init();
	scanf("%u",&t);
	while(t--)
	{
		scanf("%u",&n);
		uint res=1;
		for(uint i=1;i<=MAXPOW;++i)
		{
			uint left=0,mid,right=MAXTOT,r=0;
			while(left<=right)
			{
				mid=(left+right)>>1;
				if(QuickPow(p[mid],i)<=n)
				{
					r=mid;
					left=mid+1;
				}
				else right=mid-1;
			}
			res*=pro[r];
		}
		printf("Case %d: %u
",++CASE,res);
	}
	return 0;
}