CF236 B. Easy Number Challenge【求约数个数】

B. Easy Number Challenge

time limit per test

2 seconds

memory limit per test

256 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 ≤ 100).

Output

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

Examples

input

2 2 2

output

20

input

5 6 7

output

1520

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.

代码如下

#include<stdio.h>const int MOD=1073741824;const int MAX=1e6;int a[MAX+10];int pp()//判断一个数约数的个数{   a[1]=1;    for(int i=2; i<=MAX; i++)    {   int q=i;        a[i]=1;        for(int j=2; j*j<=q; j++)//提高效率          if(q%j==0)            {  int s=1;                while(q%j==0) s++,q/=j;                a[i]*=s;            }           if(q>1)//当它的约数中有素数时            a[i]*=2;    }}int main(){    pp();    int x,b,c,i,j,k;    long long sum;    while(~scanf("%d%d%d",&x,&b,&c))    {        sum=0;        for(i=1; i<=x; i++)            for(j=1; j<=b; j++)                for(k=1; k<=c; k++)                    sum=(sum+a[i*j*k]%MOD)%MOD;        printf("%I64d\n",sum);    }}