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.

InputInput 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).
OutputFor 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 Input51052001520Sample OutputCase 1: 2520Case 2: 60Case 3: 2300527488Case 4: 360360Case 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 long
using 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<c
4000
nt&&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;
}