HDU 5139 Formula (bestcoder Round #21)

FormulaTime limit:4000/2000 MS (java/others) Memory L imit:32768/32768 K (java/others)
Total submission (s): 306 Accepted Submission (s): 128


Problem Description F (N)=(∏ i = 1 n 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

The recursive relationship is quite simple. official:

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.

My solution.

Use an array to save hundreds of results, and then use the number group to save the result of a whole hundred factorial. In this way, each query can be divided by 100 to find the corresponding whole hundred results, and then a maximum of 99 recursion can produce results.

Code:

#include <iostream> #include <cstdio> #include <algorithm>using namespace Std;const int mod= 1000000007;long long A[100010];long long B[100010];int main () {    long long help=1;    Long long ans0=1,ans1=1;    A[0]=1;    B[0]=1;    for (int i=1;i<=10000000;i++)    {        ans1= (ans0*help)%mod;        Help = help* (i+1)%mod;        if (i%100==0)        {            a[i/100]=ans1;            b[i/100]=help;        }        ans0=ans1;    }    int n;    while (~SCANF ("%d", &n))    {        int temp=n/100;       printf ("%i64d\n", A[temp]);        ANS0=ANS1=A[TEMP];        HELP=B[TEMP];        for (int i=temp*100+1;i<=n;i++)       {        ans1= (ans0*help)%mod;        Help = help* (i+1)%mod;        ans0=ans1;       }       printf ("%i64d\n", ans1);    }    return 0;}

HDU 5139 Formula (bestcoder Round #21)