3項係数の約数について。

```import Control.Monad
import Data.Array.Unboxed
import Data.List
n = 2*10^4
d = 10
main = print p154
factor p x | mod x p ==  = succ.factor p \$ div x p
| otherwise =
factorMemo p = listArray (,n).scanl (+) .map (factor p) \$ [1..n] :: UArray Int Int
p154 = foldl'(+) \$
do a <- [..div n 3]
b <- [a..div (n-a) 2]
let c  = n-a-b
let d2 = sum.map (factor2!)\$ [a,b,c]
d5 = sum.map (factor5!)\$ [a,b,c]
guard\$ n2 - d2 >= d && n5 - d5 >= d
return\$ if a==b||b==c then 3 else 6
where n2 = factor2! n
n5 = factor5! n
factor2 = factorMemo 2
factor5 = factorMemo 5
```

```public class P154{
static int n=2*(int)Math.pow(10,5),d=12;
public static int factor(int p,int x){
return x%p==? 1+factor(p,x/p): ;
}
public static void main(String[] args){
long count=;
int[] f2 = new int[n+1],f5 = new int[n+1];
for(int i=1; i<n+1; f2[i]=f2[i-1]+factor(2,i++));
for(int i=1; i<n+1; f5[i]=f5[i-1]+factor(5,i++));
for(int a=, n2=f2[n], n5=f5[n]; a<=n/3; a++)
for(int b=a; b<=(n-a)/2; b++){
int c=n-a-b, d2=f2[a]+f2[b]+f2, d5=f5[a]+f5[b]+f5;
if(n5-d5>=d&&n2-d2>=d)count+= a==b||b==c? 3: 6;
}
System.out.println(count);
}
}
```

c++で

```#include <iostream>
using namespace std;
int n=200000,d=12;
int factor(int p,int x){
return x%p==? 1+factor(p,x/p): ;
}
int main(){
long count=;
int f2[n+1],f5[n+1];
for(int i=1; i<n+1; i++)f2[i]=f2[i-1]+factor(2,i);
for(int i=1; i<n+1; i++)f5[i]=f5[i-1]+factor(5,i);
for(int a=, n2=f2[n], n5=f5[n]; a<=n/3; a++)
for(int b=a; b<=(n-a)/2; b++){
int c=n-a-b, d2=f2[a]+f2[b]+f2, d5=f5[a]+f5[b]+f5;
if(n5-d5>=d&&n2-d2>=d)count+= a==b||b==c? 3: 6;
}
cout << count << endl;
}
```