HDU 2654 Become A Hero (欧拉函数)

Description：

Lemon 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.

Input

First line contains an integer $T(1 <= T <= 1000000)$ indicates the number of test case. Then $T$ line follows, each line contains an integer $n (1 <= n <= 2000000)$.

Output

For each data print one line, the Answer($n$).

Sample Input

1
1

Sample Output

0

题意：

求 $1 ~ n$ 中与 $n$

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9')
  {
    if (ch == '-')
      sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
  {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
  if (a > 9)
    Out(a / 10);
  putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
  return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
  return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
  ll res = 1, t = m;
  while (k)
  {
    if (k & 1)
      res = res * t % mod;
    t = t * t % mod;
    k >>= 1;
  }
  return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
  return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
  if (b == 0)
  {
    x = 1;
    y = 0;
    return a;
  }
  int g = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
  int d, x, y;
  d = exgcd(a, p, x, y);
  if (d == 1)
    return (x % p + p) % p;
  else
    return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
  int M = 1, y, x = 0;
  for (int i = 0; i < n; ++i) //算出它们累乘的结果
    M *= a[i];
  for (int i = 0; i < n; ++i)
  {
    int w = M / a[i];
    int tx = 0;
    int t = exgcd(w, a[i], tx, y); //计算逆元
    x = (x + w * (b[i] / t) * x) % M;
  }
  return (x + M) % M;
}

const int N = 2000010;
int phi[N];
void oula()
{
  phi[1] = 1;
  for (int i = 2; i < N; i++)
  {
    phi[i] = i;
  }
  for (int i = 2; i < N; i++)
  {
    if (phi[i] == i)
    {
      for (int j = i; j < N; j += i)
      {
        phi[j] = phi[j] / i * (i - 1);
      }
    }
  }
}

int main()
{
  oula();
  int t;
  sd(t);
  while (t--)
  {
    int n;
    sd(n);
    int ans = n - phi[n];
    pd(ans);
  }
  return 0;
}