codeforces236B Easy Number Challenge

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).

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.

分析：求一个数的约数有几个

#include<cstdio>#include<cstring>const int N=1000001;const int M=1<<30;int d[N],p[N];bool f[N];void get_prime(int n){int i,j,cnt=0;f[0]=f[1]=1;for(i=2;i<n;i++){if(!f[i])p[cnt++]=i;for(j=0;j<cnt&&p[j]*i<N;j++){f[i*p[j]]=1;if(i%p[j]==0)break;}}}void get_prime1(int n){int i,j,cnt=0;f[0]=f[1]=1;for(i=2;i*i<n;i++){if(f[i])continue;for(j=i;j*j<n;j++)f[i*j]=1;}for(i=2;i<N;i++)if(!f[i])p[cnt++]=i;}int count(int n){if(d[n])return d[n];int i=0,a=n,cnt;d[n]=1;while(a!=1){cnt=0;while(a%p[i]==0){cnt++;a/=p[i];}if(cnt)d[n]*=(1+cnt);i++;}return d[n];}int main(){int a,b,c,i,j,k,ans;get_prime(N);while(~scanf("%d%d%d",&a,&b,&c)){ans=0;for(i=1;i<=a;i++)for(j=1;j<=b;j++)for(k=1;k<=c;k++){ans+=count(i*j*k);if(ans>=M)ans-=M;}printf("%d\n",ans);}return 0;}

方法二：

#include<cstdio>#include<cstring>const int N=1000001;const int M=1<<30;int d[N];int main(){int a,b,c,i,j,k,ans,J;while(~scanf("%d%d%d",&a,&b,&c)){ans=0;J=a*b*c;memset(d,0,sizeof(d));for(i=1;i<=J;i++)for(j=1;i*j<=J;j++)d[i*j]++;for(i=1;i<=a;i++)for(j=1;j<=b;j++)for(k=1;k<=c;k++){ans+=d[i*j*k];if(ans>=M)ans-=M;}printf("%d\n",ans);}return 0;}