uva 10820 欧拉函数

这里学习一下欧拉函数。欧拉函数是解决类似这样的问题模型：给出n的唯一分解式n=p1^a1*p2^a2..................pk^ak求1,2........n的互素的数的个数。（证明以后补上，囧。还没有太理解，以后回来填个坑）

这是模板：

LL phi[N],ans[N];void phi_table(){memset(phi, 0, sizeof(phi));phi[1] = 1;for (LL i = 2; i <= N; i++)if (!phi[i])for (LL j = i; j <= N; j += i){if (!phi[j])phi[j] = j;phi[j] = phi[j] / i*(i - 1);}}

这一题题意的本质是让你求出n以内有多少个二元满足（1<=x,y=<n）&&(x,y互质)，求出n以内的欧拉函数值得和即f（N）=phi[2]+.............phi[n],输出2*f（n）+1，为啥是2*f(n),

存在（a，b）互质二元组则即存在（b，a），加上1，是还有（1,1）这个特殊情况

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 50009#define LL long long LL phi[N],ans[N];void phi_table(){memset(phi, 0, sizeof(phi));phi[1] = 1;for (LL i = 2; i <= N; i++)if (!phi[i])for (LL j = i; j <= N; j += i){if (!phi[j])phi[j] = j;phi[j] = phi[j] / i*(i - 1);}}void solve(){ans[1] = 0;for (int i = 2; i <= N; i++){ans[i] = ans[i - 1] + phi[i];}}int main(){#ifdef CDZSCfreopen("i.txt", "r",stdin);#endifphi_table();solve();int n;while (~scanf("%d", &n) && n){printf("%lld\n", 2 * ans[n] + 1);}return 0;}

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
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
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