CRB and Candies (LCM (C (n,0): C (n,n) =LCM (,,, n+1)/(n+1)) hdu5407

CRB and Candies

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 358 Accepted Submission (s): 160

problem DescriptionCRB has  N  Different candies. He is going to eat  K  Candies.
He wonders how many combinations he can select.
Can you answer he question for all  K (0≤  K  ≤  N )?
CRB is too hungry to check all the your answers one by one, so he only asks least common multiple (LCM) of all answers.

Input

There is multiple test cases. The first line of input contains an integer  T , indicating the number of test cases. For all test case there are one line containing a single integer  N .
1≤  T  ≤300
1≤  N  ≤  10^ 6

Output

For each test case, output a single INTEGER–LCM modulo 1000000007 ( 10^9+7 ).

Sample Input

512345

Sample Output

1231210

Author

KUT (DPRK)

Source

Multi-university Training Contest 10

Solving:

Reprint Please specify the Source: Looking for Children & stars   

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5407

LCM (c (n,0), C (n,1), C (n,2),,,, C (n,n)) =LCM (,,,, n,n+1)/(n+1)//multiplication Inverse! (Division + Modulo)//f (1) =1//if n=p^k then f (n) =f (n-1) *p else f (n) =f (n-1) #include <stdio.h> #include <string.h> #define MoD 1000000007#define ll __int64const ll Maxv=1e6+5;bool Isnp[maxv]={false};    ll prime[maxv],pnum;//Prime Array, number of elements ll cas,f[maxv]={0};void get_prime ()//prime table {pnum=0; LL i,j;    memset (isnp,0,sizeof (ISNP));    Isnp[0]=isnp[1]=true;        for (i=2; i<maxv; i++) {if (!isnp[i]) {prime[pnum]=i;pnum++;}            for (j=0; j<pnum&&prime[j]*i<maxv; J + +) {isnp[i*prime[j]]=true;        if (i%prime[j]==0) break;        }} for (i=0;i<pnum;i++) {for (J=prime[i];j<maxv;j*=prime[i]) {f[j]=prime[i];    }}}void init () {get_prime ();    F[1]=1;        for (LL i=2;i<maxv;i++) {if (f[i]) f[i]=f[i]*f[i-1]%mod;    else f[i]=f[i-1];   }//for (LL i=2;i<100;i++)//{//printf ("I=%i64d\t f=%i64d\n", i,f[i]);//     if (i%10==0) printf ("\ n"),//}}void EXGCD (LL a,ll b,ll &d,ll &x,ll &y) {if (!b) {d=a;x=1;y=0;}        else {EXGCD (b,a%b,d,y,x);    Y-=x* (A/b);    }}int Main () {int T;    LL N;    Init ();    scanf ("%d", &t);        while (t--) {scanf ("%i64d", &n);        LL x,y,d;        EXGCD (n+1,mod,d,x,y);//ax = 1 (mod m) if (d==1) {x= (x%mod+mod)%mod;}    printf ("%i64d\n", f[n+1]*x%mod); } return 0;}

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CRB and Candies (LCM (C (n,0): C (n,n) =LCM (,,, n+1)/(n+1)) hdu5407