Discription
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).
Example
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.
我是直接暴力合并a和b，然后设 to(i) 为有多少对有序对(x,y) 满足 1<=x<=a 且 1<=y<=b 且 x*y==i。
然后式子就是 Σ(i=1 to a*b) to(i) Σ(j=1 to c) d(i*j)
这个用约数个数函数的基本式子就可以化简，最后可以 用不到 O(a*b*log(a*b)) 的时间计算出答案。
所以a,b就取三个数中最小的两个就行了。
#include<bits/stdc++.h>
#define ll long long
#define maxn 4000000
using namespace std;
const int ha=1<<30;
int a,b,c,miu[maxn+5];
int zs[maxn/5],t=0,D,low[maxn+5];
int d[maxn+5],to[maxn+5],ans=0;
bool v[maxn+5];
inline void init(){
for(int i=1;i<=a;i++)
for(int j=1;j<=b;j++) to[i*j]++;
miu[1]=d[1]=low[1]=1;
for(int i=2;i<=maxn;i++){
if(!v[i]) zs[++t]=i,miu[i]=-1,low[i]=i,d[i]=2;
for(int j=1,u;j<=t&&(u=zs[j]*i)<=maxn;j++){
v[u]=1;
if(!(i%zs[j])){
low[u]=low[i]*zs[j];
if(low[i]==i) d[u]=d[i]+1;
else d[u]=d[low[u]]*d[i/low[i]];
break;
}
low[u]=zs[j],d[u]=d[i]<<1,miu[u]=-miu[i];
}
}
for(int i=1;i<=maxn;i++) d[i]+=d[i-1];
}
inline int add(int x,int y){
x+=y;
return x>=ha?x-ha:x;
}
inline void calc(){
for(int i=1,sum;i<=c;i++) if(miu[i]){
sum=0;
for(int j=i;j<=D;j+=i) sum=add(sum,to[j]*(ll)(d[j/i]-d[j/i-1])%ha);
sum=sum*(ll)d[c/i]%ha;
if(miu[i]==1) ans=add(ans,sum);
else ans=add(ans,ha-sum);
}
printf("%d
",ans);
}
int main(){
scanf("%d%d%d",&a,&b,&c);
if(a>b) swap(a,b);
if(a>c) swap(a,c);
if(b>c) swap(b,c);
D=a*b;
init();
calc();
return 0;
}