lightoj-1289(数论+素数打表)

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 modulo2³².

For each case, print the case number and lcm(1, 2, 3, ..., n). As the result can be very big, print the result modulo2³².

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

题目题意：就是让我们求lcm(1,2,,,,n).

题目分析: 这里我们求得时候肯定不能暴力求，肯定超时，这里我们要用到一个结论:

lcm(1,2,,,,,,n+1)=lcm(1,2,,,,,,,n)*p   （这里的p是指素数，n+1=p^k ,如果n+1可以写成这样的形式，那么就得到那样的递推式)

                         =lcm(1,,2,,n)      others

这个题目还要用到素数的打表的另一种写法，用位图（我表示不是很懂，但是它的两个函数给出后，就可以与以前的标记数组vis一样用了）

代码如下:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define SHIFT 5#define ll long longusing namespace std;ll inf;const int maxn=1e8+2;int prime[6000000+10],cnt;unsigned int sum[6000000+10];unsigned int vis[maxn>>SHIFT];void SetBit(int x){    vis[x>>SHIFT]|=1<<(x&((1<<SHIFT)-1));}bool GetBit(int x){    return vis[x>>SHIFT]&(1<<(x&((1<<SHIFT)-1)));}void get_prime(){    for (int i=2;i<maxn;i++) {        if (!GetBit(i)) prime[cnt++]=i;//这里就相当于vis[i]    for (int j=0;j<cnt&&i*prime[j]<maxn;j++) {         SetBit(i*prime[j]);//相当于vis[i*prime[j]]=false        if (i%prime[j]==0) break;    }   }}void init()//把质数先乘起来，因为质数肯定满足{    sum[0]=prime[0];    for (int i=1;i<cnt;i++)        sum[i]=sum[i-1]*prime[i];}unsigned int solve(int n){    int x=upper_bound(prime,prime+cnt,n)-prime-1;//找到n或者比它小一点的位置    unsigned int ans=sum[x];    for (int i=0;i<cnt&&prime[i]*prime[i]<=n;i++) {        int cur=prime[i];        int temp=prime[i]*prime[i];        while (temp/cur==prime[i]&&temp<=n) {//找到每一个质数prime[i]，它在n之内有多少个prime[i]^k，有一个就得乘一个prime[i]            cur*=prime[i];            temp*=prime[i];        }        ans=ans*(cur/prime[i]);    }    return ans;}int main(){get_prime();init();inf=pow(2,32);int t;scanf("%d",&t);for (int icase=1;icase<=t;icase++) {        int n;        scanf("%d",&n);printf("Case %d: %u\n",icase,solve(n)%inf);}return 0;}