唯一分解定理+n個數的LCM

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

5

10

5

200

15

20

Sample Output

Case 1: 2520

Case 2: 60

Case 3: 2300527488

Case 4: 360360

Case 5: 232792560

參考博客：https://blog.csdn.net/whyorwhnt/article/details/9397289

就是先累乘出素數的前綴和，然後找到比n小的第一個素數。然後就枚舉那個素數（<n）然後看他的次方是否滿足。

然後用ASN乘起來這個數。

#include<bits/stdc++.h>
using namespace std;

const int N=100000066;

int visit[N/32+50];
unsigned int data[5800000];
int prime[5800000],np=0;

void Prime ()   //篩素數,數組從0開始
{
	prime[0]=data[0]=2;
	np=1;
	for (int i=3;i<N;i+=2)   //掃所有奇數
		if (!(visit[i/32] & (1 << ((i/2)%16))))
		{
			prime[np]=i;
			data[np]=data[np-1]*i;  //預處理
			np++;
			for (int j=3*i;j<N;j+=2*i)  //改成i*i會超int範圍
				visit[j/32] |= (1 << ((j/2)%16));
		}
}

unsigned int Deal (int n)
{
	int p=upper_bound (prime, prime+np, n)-prime-1;  //定位比n小的第一個素數
	unsigned int ans = data[p];
	//cout<<ans<<"ans"<<endl;
	for (int i=0; i<np && prime[i]*prime[i] <= n; i++)//此時prime[i]最多10^4
	{//掃所有素數的整數次冪
		int mul = prime[i];//對於每一個素數
		int tmp = prime[i] * prime[i];//它的平方
		while (tmp/mul == prime[i] && tmp<=n) //防止int越界
		{
			tmp *= prime[i];//
			mul *= prime[i];
		}
		ans *= (mul/prime[i]);
		//這裏是看幾的次方爲一個素數，再乘上就可以了。
		//cout<<ans<<"ans111111"<<endl;
	}
	return ans;
}

int main ()
{
	int T,n;
	scanf("%d",&T);
	Prime ();
	for (int Cas=1;Cas<=T;Cas++)
	{
		scanf ("%d",&n);
		printf ("Case %d: %u\n",Cas,Deal(n));
	}
	return 0;
}