http://codeforces.com/contest/1114/problem/C
C. Trailing Loves (or L'oeufs?)
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
The number "zero" is called "love" (or "l'oeuf" to be precise, literally means "egg" in French), for example when denoting the zero score in a game of tennis.
Aki is fond of numbers, especially those with trailing zeros. For example, the number 92009200 has two trailing zeros. Aki thinks the more trailing zero digits a number has, the prettier it is.
However, Aki believes, that the number of trailing zeros of a number is not static, but depends on the base (radix) it is represented in. Thus, he considers a few scenarios with some numbers and bases. And now, since the numbers he used become quite bizarre, he asks you to help him to calculate the beauty of these numbers.
Given two integers nn and bb (in decimal notation), your task is to calculate the number of trailing zero digits in the bb-ary (in the base/radix of bb) representation of n!n! (factorial of nn).
Input
The only line of the input contains two integers nn and bb (1≤n≤10181≤n≤1018, 2≤b≤10122≤b≤1012).
Output
Print an only integer — the number of trailing zero digits in the bb-ary representation of n!n!
Examples
input
Copy
6 9
output
Copy
1
input
Copy
38 11
output
Copy
3
input
Copy
5 2
output
Copy
3
input
Copy
5 10
output
Copy
1
Note
In the first example, 6!(10)=720(10)=880(9)6!(10)=720(10)=880(9).
In the third and fourth example, 5!(10)=120(10)=1111000(2)5!(10)=120(10)=1111000(2).
The representation of the number xx in the bb-ary base is d1,d2,…,dkd1,d2,…,dk if x=d1bk−1+d2bk−2+…+dkb0x=d1bk−1+d2bk−2+…+dkb0, where didi are integers and 0≤di≤b−10≤di≤b−1. For example, the number 720720 from the first example is represented as 880(9)880(9) since 720=8⋅92+8⋅9+0⋅1720=8⋅92+8⋅9+0⋅1.
You can read more about bases here.
开始超时了 ，一开始的想法是看这个n!里面有b，那么末尾就有多少个0。然后搞了一个从2到n的循环，可以整除b，就ans++。大于b，就对b取模。
正解：将b分解因子，比如b可以分解成2个3，4个7。那么ans=min(n!中3的个数/2，n!中7的个数/4);
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define res register int
#define inf 0x3f3f3f3f
const int maxn=1e5;
ll a[maxn],num[maxn];
ll cal(ll y,ll x)
{
ll ress=0;
while(x){
ress+=x/y;
x/=y;
}
return ress;
}
int main()
{
ll n,b,temp,cnt=0;
cin>>n>>b;
for(ll i=2;i*i<=b;i++){
if(0==b%i){
a[++cnt]=i;
while(0==b%i){
num[cnt]++;
b/=i;
}
}
}
if(b>1) num[++cnt]=1,a[cnt]=b;
ll ans=0;
for(ll i=1;i<=cnt;i++){
if(1==i) ans=cal(a[i],n)/num[i];
else ans=min(ans,(ll)(cal(a[i],n)/num[i]));
}
cout<<ans;
return 0;
}