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).
Sample test(s)
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.
解题说明：此题是求一个数因子的个数，打表求出每个数因子的个数，相加
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include<set>
#include <algorithm>
using namespace std;
int x[1000003];
int main()
{
int a,b,c;
int i,j,k,s=0;
scanf("%d %d %d",&a,&b,&c);
for(i=2;i<=1000002;++i)
{
for( j=i;j<=1000002;j+=i)
{
++x[j];
}
}
for(i=1;i<=a;++i)
{
for(j=1;j<=b;++j)
{
for( k=1;k<=c;++k)
{
s=(s+1+x[i*j*k])%1073741824;
}
}
}
printf("%d\n",s);
return 0;
}