【UVA 10820】Send a Table（欧拉函数）

Description

---

When participating in programming contests, you sometimes face the following problem: You know

how to calcutale the output for the given input values, but your algorithm is way too slow to ever

pass the time limit. However hard you try, you just can’t discover the proper break-off conditions that

would bring down the number of iterations to within acceptable limits.

Now if the range of input values is not too big, there is a way out of this. Let your PC rattle for half

an hour and produce a table of answers for all possible input values, encode this table into a program,

submit it to the judge, et voila: Accepted in 0.000 seconds! (Some would argue that this is cheating,

but remember: In love and programming contests everything is permitted).

Faced with this problem during one programming contest, Jimmy decided to apply such a ’technique’.

But however hard he tried, he wasn’t able to squeeze all his pre-calculated values into a program

small enough to pass the judge. The situation looked hopeless, until he discovered the following property

regarding the answers: the answers where calculated from two integers, but whenever the two

input values had a common factor, the answer could be easily derived from the answer for which the

input values were divided by that factor. To put it in other words:

Say Jimmy had to calculate a function Answer(x, y) where x and y are both integers in the range

[1, N]. When he knows Answer(x, y), he can easily derive Answer(k ∗ x, k ∗ y), where k is any integer

from it by applying some simple calculations involving Answer(x, y) and k.

For example if N = 4, he only needs to know the answers for 11 out of the 16 possible input value

combinations: Answer(1, 1), Answer(1, 2), Answer(2, 1), Answer(1, 3), Answer(2, 3), Answer(3, 2),

Answer(3, 1), Answer(1, 4), Answer(3, 4), Answer(4, 3) and Answer(4, 1). The other 5 can be derived

from them (Answer(2, 2), Answer(3, 3) and Answer(4, 4) from Answer(1, 1), Answer(2, 4) from

Answer(1, 2), and Answer(4, 2) from Answer(2, 1)). Note that the function Answer is not symmetric,

so Answer(3, 2) can not be derived from Answer(2, 3).

Now what we want you to do is: for any values of N from 1 upto and including 50000, give the

number of function Jimmy has to pre-calculate.

Input

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The input file contains at most 600 lines of inputs. Each line contains an integer less than 50001 which

indicates the value of N. Input is terminated by a line which contains a zero. This line should not be

processed.

Output

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For each line of input produce one line of output. This line contains an integer which indicates how

many values Jimmy has to pre-calculate for a certain value of N.

Sample Input

2
5
0

Sample Output

3
19

题解

---

问小于n，且两元素互素的二元组有多少个，答案为

\[(2\cdot\sum_{i=1}^{n}phi(i))-1$$（二元组可以倒置,(1,1)除外）

###参考代码
```C++
#include <queue>
#include <cmath>
#include <cstdio>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define inf 1000000000
#define PI acos(-1)
#define mem(a,x) memset(a,x,sizeof(a))
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void Out(ll a){
if(a<0) putchar('-'),a=-a;
if(a>=10) Out(a/10);
putchar(a%10+'0');
}
const int N=50005;
int phi[N];
void phi_table(int n){
for(int i=2;i<=n;i++) phi[i]=0;
phi[1]=1;
for(int i=2;i<=n;i++) if(!phi[i]){
for(int j=i;j<=n;j+=i){
if(!phi[j]) phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
}
}
int ans[N];
int main(){
phi_table(50000);
for(int i=1;i<=50000;i++) ans[i]=ans[i-1]+phi[i];
while(true){
int n=read();
if(!n) break;
Out(2*ans[n]-1);
puts("");
}
return 0;
}
```\]