1694:primorial vs LCM number theory

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1694:primorial vs LCM time limit: 4 Sec Memory limit: mb[submission [status] [discussion] Title description
Given N (2<=n<=10^14), what's the quotient of LCM (,...., 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 the LCM stands for Lowest or Least Common multiple.
Input
The first line of the input was 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.
Output
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.
Sample input
10
2
3
4
5
6
7
8
9
10
1000
Sample output
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 10:744,593,350

Test Instructions: Find out LCM (1~n)/(prime number product within years)

Ideas: The product of the highest power of the energy of each prime number within the LCM (1~n) = n

LCM (1,2,3,4,,, 49) = 2^5 * 3^3 * 5^2 * 7^2 * 11 *13 ...

Find LCM (1~49) except for 49 primes = 2^4 * 3^3 * 5 * 7

This only needs to calculate the prime number within 1E7 and then take the highest of each prime power-1, you can find out the situation of n=1e14 but still will be T

Since the answer to a continuous n is the same for all to simulate the process (the conclusion here is that the value of the LCM is not changed when a new prime number of the highest power is present)

2 1

3 1

4 2

8 4

9 12

16 24

25 120

Prime^num anw[on a]*prime

below to analyze and analyze

Think about what this is about, because each time a new prime is given the highest power to change the LCM.

So every time the answer is updated, it's here at Prime^num.

And then in the linear sieve there is a record value and then two points output the answer

Code:

#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 1000000007int t,cas;vector< pair&    lt;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;}

1694:primorial vs LCM number theory