链接：戳这里
1694: Primorial vs LCM
时间限制: 4 Sec 内存限制: 128 MB
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题目描述
Given N (2<=N<=10^14), what is the quotient of LCM(1,2,3,....,N) divided by multiple of all primes up
to N. As the result might be too big, output it‘s modulo by 1000000007.
For example, when N=5, the result is LCM(1,2,3,4,5)/(2*3*5)=60/30=2.
Note that LCM stands for Lowest or Least Common Multiple.
输入
The first line of the input is T(T ≤ 50000), then T test cases follows in next T lines. Each line
contains an integer N (2 ≤ N ≤ 100000000000000 or 10^14). The meaning of N is given in the
problem statement.
输出
For each test case print a line in “Case x: S” format where x is case number and S is the
quotient modulo by 1000000007.
样例输入
10
2
3
4
5
6
7
8
9
10
1000
样例输出
Case 1: 1
Case 2: 1
Case 3: 2
Case 4: 2
Case 5: 2
Case 6: 2
Case 7: 4
Case 8: 12
Case 9: 12
Case 10: 744593350
题意：求出LCM(1~n)/(年以内的素数乘积)
思路：LCM(1~n) = n以内的每个素数的能取的最高次幂的乘积
LCM(1,2,3,4,,,49) = 2^5 * 3^3 * 5^2 * 7^2 * 11 *13...
发现LCM(1~49) 除一下 49以内的素数 = 2^4 * 3^3 * 5 * 7
这样只需要计算1e7以内的素数然后取每个素数的最高次幂-1,就可以求出n=1e14的情况 但是还是会T
由于一段连续的n的答案是一样的 所有来模拟一下过程 (这里的结论是没有当出现一个新的最高次幂的素数的时候才会改变LCM的值
2 1
3 1
4 2
8 4
9 12
16 24
25 120
prime^num anw[上一项]*prime
下面来分析分析
想一下这是什么原因呢 因为每次只有得到新的prime得最高次幂才会对LCM产生变化
所以每次的答案更新也就在prime^num这里
哎 没意思 然后在线性筛那里记录值 然后二分输出答案
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 10000005
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const ll INF=1e14+5;
#define mod 1000000007
int T,cas;
vector< pair<ll,int> > V;
vector<ll> anw;
int vis[MAX],prime[MAX];
void init(){
int cnt=0;
V.push_back(make_pair(1LL,1));
for(int i=2;i<=MAX;i++){
if(!vis[i]){
prime[cnt++]=i;
for(double j=(double)i*i;j<INF;j*=i){
V.push_back(make_pair((ll)j,i));
}
}
for(int j=0;j<cnt;j++){
if((ll)i*prime[j]>MAX) break;
vis[i*prime[j]]=1;
if(i%prime[j]==0) break;
}
}
sort(V.begin(),V.end());
anw.push_back(1LL);
for(int i=1;i<V.size();i++){
anw.push_back((ll)V[i].second*anw[i-1]%mod);
}
}
ll n;
int main(){
init();
scanf("%d",&T);
for(int cas=1;cas<=T;cas++){
scanf("%lld",&n);
int l=0,r=V.size()-1,mid,ans=0;
while(l<r){
mid=(l+r+1)/2;
if(V[mid].first<=n) {
l=mid;
ans=mid;
}
else r=mid-1;
}
printf("Case %d: %lld\n",cas,anw[r]%mod);
}
return 0;
}