Hdoj 5139 Formula offline

Find the Law f (1) =1f (2) =1*1*2= (1) * (1*2) =1!*2!f (3) =1*1*1*2*2*3= (1) * (1*2) * (1*2*3) =1!*2!*3! formula can be simplified to
    
     F  (N)=  ∏i=1n  (N!)%M  OD
    , directly hit the table does not, will be ultra-memory, data can be processed offline. After a good order from small to large violence. clogc+10000000, C is the number of case.

Formula Time limit:4000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 714 Accepted Submission (s): 255


Problem Description F (N)=( ∏i=1n in? I + 1 )%1000000007
You were expected to write a program to calculate f (n) when a certain n was given.
Inputmulti test Cases (about 100000), every case contains an integer n-a single line.
Please process to the end of file.

[Technical specification]
1≤n≤10000000
Outputfor each n,output f (n) in a.
Sample Input

2100

Sample Output

2148277692

Sourcebestcoder Round #21

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Vector>using namespace Std;typedef Long Long int ll;const ll maxn=10001000;const ll Mod=1000000007;int a,id;struct QUE{ int x,id; LL ans;} Que[maxn];bool cmp1 (que a,que b) {return a.x<b.x;} BOOL Cmp2 (QUE A,que b) {return a.id<b.id;} LL Now,nowans; LL Nowjiecheng; ll Jiec (ll x) {for (int i=now+1;i<=x;i++) {nowjiecheng= (nowjiecheng*i)%mod;} Now=x;return Nowjiecheng;} int main () {while (scanf ("%d", &a)!=eof) {que[id].x=a; que[id].id=id;id++;} Sort (QUE,QUE+ID,CMP1); now=1,nowans=1,nowjiecheng=1;for (int i=0;i<id;i++) {if (Que[i].x==now) {Que[i].ans=nowans ; }else if (que[i].x>now) {LL temp=1;for (int j=now+1;j<=que[i].x;j++) {temp= (Temp*jiec (j))%mod;} Nowans= (Nowans*temp)%mod;que[i].ans=nowans;}} Sort (QUE,QUE+ID,CMP2); for (int i=0;i<id;i++) {Cout<<que[i].ans<<endl;} return 0;}

Hdoj 5139 Formula offline