Codeforces 235E. Number Challenge DP

dp(a,b,c,p) = sigma ( dp(a/p^i,b/p^j,c/p^k) * ( 1+i+j+k) )

表示用小于等于p的素数去分解的结果有多少个

E. Number Challenge

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Sample test(s)

input

2 2 2

output

20

input

4 4 4

output

328

input

10 10 10

output

11536

Note

For the first example.

• d(1·1·1) = d(1) = 1;
• d(1·1·2) = d(2) = 2;
• d(1·2·1) = d(2) = 2;
• d(1·2·2) = d(4) = 3;
• d(2·1·1) = d(2) = 2;
• d(2·1·2) = d(4) = 3;
• d(2·2·1) = d(4) = 3;
• d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

import java.util.*;

public class CF235E {

	class Triple {
		
		Triple(){}
		
		Triple(int _x,int _y,int _z) {
			this.x=_x; this.y=_y; this.z=_z;
			this.sort();
		}
		
		public int x,y,z;
		
		void sort() {
			if(this.z();
	}
	
	
	
	long gao(int deep,Triple tri) {
		
		if(deep==pn) return 1;
		
		if(map[deep].get(tri)!=null) return (long) map[deep].get(tri);
		long ret=0;
		
		int p=primes[deep];
		
		for(int x=tri.x,i=0;x!=0;x/=p,i++) {
			for(int y=tri.y,j=0;y!=0;y/=p,j++) {
				for(int z=tri.z,k=0;z!=0;z/=p,k++) {
					ret+=gao(deep+1,new Triple(x,y,z))*(i+j+k+1)%mod;
					if(ret>=mod) {
						ret-=mod;
					}
				}
			}
		}
		map[deep].put(tri, ret);
		return ret;
	}
	
	CF235E(){
		init();		
		Scanner in = new Scanner(System.in);
		a=in.nextInt(); b=in.nextInt(); c=in.nextInt();
		System.out.println(gao(0,new Triple((int)a,(int)b,(int)c)));
	}
	
	public static void main(String[] args) {
		new CF235E();
	}
}