Test instructions

Enter N to find the least common multiple of all the combined numbers of C (n,0) to C (N,n).

Input:

The first line inputs an integer t, which indicates a common T Group test Sample.

Each set of test sample contains a positive integer n (1<=n<=1e6).

Output:

Output results (mod 1e9+7).

Feeling pretty sick, from the beginning of the game I was writing this problem, the game is over or not written out ...

During the period found the inverse, least common multiple, combined number of various formulas, but blew an afternoon tle.

The game is over, the puzzle tells me, formula seconds kill law ...

But the formula can not understand, fortunately there is a group of giant commentary, so some understand, but still need to continue to think to understand.

Exercises

Set Ans[i] Represents the least common multiple of all combinations of I.

Set F[i] Represents the least common multiple of positive integers from 1 to I.

Then get the formula--ans[i] = f[i+1]/i from f[i] to ans[i]. Anss[i-1] = (f[i]*inv[i])%mod;

Then get the formula for F[i]--

if (i = = p^k) F[i] = f[i-1]*p;

else f[i] = f[i-1];

The next thing to do is to find those i = = p^k.

Problem Solving steps:

1. Number of prime tables;

2. Looking for i = = P^k by the prime number table;

3. f[1] = 1, hit f[] array table and ans[] array table;

4. Input data;

5. Output the answer based on the input data and ans[] table.

I think there are other ways, since the operating time of the AC program is from 15ms to 900+ms. More ways to keep looking.

My (according to the puzzle) code--

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5 using namespacestd;6 7 #defineLL Long Long8 Const intN =1000010;9 Const intMod =1000000007;Ten One intN, t; A LL ans, mid; - LL Inv[n]; - LL Anss[n]; the BOOLSu2[n]; - intSu[n], su1[n]; -LL F[n];//F[i] Represents the least common multiple of 1~i - + - voidtable () + { Ainv[1] =1; at for(inti =2; i < N; i++)//to find the inverse element of I - { -Inv[i] = inv[mod%i]* (mod-mod/i)%Mod; - } - -memset (SU2,1,sizeof(SU2)); inmemset (SU1,0,sizeof(SU1)); -su2[0] = su2[1] =0;//the number of primes within n is calculated tosu2[2] =1; + for(inti =3; i < N; i++) Su2[i] = i%2==0?0:1; - for(inti =3; I <= sqrt (n1.0); i++) the { * if(Su2[i]) $ {Panax Notoginseng for(intj = i*i; J <= N; J + =2*i) - { theSU2[J] =0; + } A } the } + intK =0; - for(inti =0; i < N; i++)//Number of prime tables within n $ { $ if(Su2[i] = =1) su[k++] =i; - } - for(inti =0; I < K; i++)//find the number that satisfies P^k, where p is prime and K is a positive integer the { -LL mid =Su[i];Wuyi while(Mid <N) the { -Su1[mid] =Su[i]; WuMid *=Su[i]; - } About } $f[1] =1;//least common multiple table of 1~i within n - for(inti =2; i < N; i++) - { - if(Su1[i]) f[i] = f[i-1]*Su1[i]; A ElseF[i] = f[i-1]; +F[i]%=Mod; theanss[i-1] = (f[i]*inv[i])%mod;//Answer table - } $ } the the intMain () the { the //freopen ("test.in", "R", stdin); - //freopen ("Test.out", "R", stdout); in table (); thescanf"%d", &t); the while(t--) About { thescanf"%d", &n); theprintf"%i64d\n", Anss[n]); the } + return 0; -}

View Code

HDU 5407 CRB and candies (combined number + least common multiple + prime table + inverse) and 2015 Summer Multi-School games 10th