UVA 11426 GCD - Extreme (II) (欧拉函数)
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Problem J
GCD Extreme (II)
Input: Standard Input
Output: Standard Output
Given the value of N, you will have to find the value of G. The definition of G is given below:
Here GCD(i,j) means the greatest common divisor of integer i and integer j.
For those who have trouble understanding summation notation, the meaning of G is given in the following code:
G=0; for(i=1;i<N;i++) for(j=i+1;j<=N;j++) { G+=gcd(i,j); } /*Here gcd() is a function that finds the greatest common divisor of the two input numbers*/ |
Input
The
input file contains at most 100 lines of inputs. Each line contains an
integer N (1<N<4000001). The meaning of N is given in the problem
statement. Input is terminated by a line containing a single zero.
Output
For
each line of input produce one line of output. This line contains the
value of G for the corresponding N. The value of G will fit in a 64-bit
signed integer.
Sample Input Output for Sample Input
10 100 200000 0
|
67 13015 143295493160
|
Problemsetter: Shahriar Manzoor
Special Thanks: SyedMonowarHossain
设dp[i]=gcd(1,i)+gcd(2,i)+……+gcd(i-1,i);
则ans[n]=dp[2]+dp[3]+……+dp[n].
由此问题已经转化成如何求dp[i]了,即需要求1到i-1所有数与i的gcd的和。
设k为满足gcd(x,i)=j且x<i的正整数的个数,则dp[i]=∑j*k;
同时,由于gcd(x,i)=j等价于gcd(x/j,i/j)=1,也就是phi[i/j];
接下来反过来求,那就不需要分解素因子了
#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn=;
int phi[maxn];
ll dp[maxn+];
ll ans[maxn+];
void phi_table()
{
phi[]=;
for(int i=;i<maxn;i++)
{
if(!phi[i])
{
for(int j=i;j<maxn;j+=i)
{
if(!phi[j])phi[j]=j;
phi[j]=phi[j]/i*(i-);
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
phi_table();
for(int i=;i<maxn;i++)
{
for(int j=i*;j<maxn;j+=i)dp[j]+=(long long)i*(long long)phi[j/i];
}
ans[]=dp[];
for(int i=;i<maxn;i++)ans[i]=ans[i-]+dp[i];
int n;
while(cin>>n&&n)
{
cout<<ans[n]<<endl;
}
return ;
}
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