Prime factor Decomposition: nefu118 (how many 0 are behind n!) + nefu119 (combined primes)

how many 0 in the back of n!?

Description

Reads a number n from the input and evaluates n. The number of the end 0.

input

The input has several lines. The first line has an integer m, indicating the number of digits that follow. Then the M-line, each row contains a positive integer n,1<=n<=1000000000 that is determined.

Output

For each data in the input row, n, the output line, whose content is N. The number of the end 0.

Sample_input

3 3 1024

Sample_output

0 253

Problem Solving Ideas:

The minimum value of the factor 2 and 5 power is obtained by the decomposition of the element factor.

Number theory knowledge:the power of prime p in n! factor decomposition is [n/p]+[n/p^2]+[n/p^3]+[n/p^4]+ ...

Reference Code:

#include 
#include 
#include 
#include #include #include
using namespace std;
const int MAXN=50000+10;
int n;
int solve (int p)
{
   int cnt=0;
   int q=p;
   while (q<=n) {
    cnt+=n/q;
    q*=p;
   }
   return cnt;
}
int main ()
{
   //freopen ("Input.txt", "R", stdin);
    int t;cin>>t;
    while (t--) {
      cin>>n;
      int ans=min (Solve (2), solve (5));
      cout<

Combined primes

Description

Xiao Ming's father came back from the outside to bring her a gift, Xiao Ming happy to run back to his room, opened a look is a very large board (very large), Xiao Ming was disappointed. But not a few days found the big board of fun. The number of non-descending paths from the starting point (0,0) to the end point (N,n) is C (2n,n), and now xiaoming randomly takes out 1 prime numbers p, and he wants to know how many times C (2n,n) is exactly divisible by P. Xiao Ming thought for a long time have not thought out, now want to ask you to help Xiao Ming solve this problem, for you should not be difficult!

input

There are multiple sets of test data. The first line is a positive integer t, which represents the number of groups of test data. The next 2 numbers for each group are the values of N and P, 1<=n,p<=1000000000 here.

Output

For each set of test data, the output line gives the number of times C (2n,n) is divisible by prime p, and when it is not divisible, the number is 0.

Sample_input

2 2 2) 2 3

Sample_output

1 1

Problem Solving Ideas:

Factor decomposition, based on the knowledge of number theory in nefu118, requires only (2n). Minus twice times the power of P (n). The power of P;

Pit in the q=q*p,q may be data is out, (p in 0-1000, 000,000, Range), big data will be error, change to a long long can.

Reference Code:

#include 
#include 
#include 
#include #include #include
using namespace std;
const int MAXN=50000+10;
int p;
int solve (int n)
{
   int cnt=0;
   Long long q=p;
   while (q<=n) {
    cnt+=n/q;
    q*=p;
   }
   return cnt;
}
int main ()
{
 //  freopen ("Input.txt", "R", stdin);
   int n;
    int t;cin>>t;
    while (t--) {
      cin>>n>>p;
      int Ans=solve (2*n) -2*solve (n);
      cout<