hdu 2654(欧拉函数)
Become A Hero
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 210 Accepted Submission(s): 57
wants to be a hero since he was a child. Recently he is reading a book
called “Where Is Hero From” written by ZTY. After reading the book,
Lemon sends a letter to ZTY. Soon he recieves a reply.
Dear Lemon,
It is my way of success. Please caculate the algorithm, and secret is behind the answer. The algorithm follows:
Int Answer(Int n)
{
.......Count = 0;
.......For (I = 1; I <= n; I++)
.......{
..............If (LCM(I, n) < n * I)
....................Count++;
.......}
.......Return Count;
}
The LCM(m, n) is the lowest common multiple of m and n.
It is easy for you, isn’t it.
Please hurry up!
ZTY
What a good chance to be a hero. Lemon can not wait any longer. Please help Lemon get the answer as soon as possible.
line contains an integer T(1 <= T <= 1000000) indicates the
number of test case. Then T line follows, each line contains an integer n
(1 <= n <= 2000000).
1
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;
typedef long long LL;
const int N = ;
int euler[N];
void getEuler()
{
memset(euler,,sizeof(euler));
euler[] = ;
for(int i = ; i <= N; i++){
if(!euler[i])
for(int j = i; j <= N; j+= i)
{
if(!euler[j])
euler[j] = j;
euler[j] = euler[j]/i*(i-);
}
}
} int main()
{
getEuler();
int tcase;
scanf("%d",&tcase);
while(tcase--){
int n;
scanf("%d",&n);
printf("%d\n",n-euler[n]);
}
return ;
}
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