直接套模板,这道题貌似单独求还快一些
解法一
#include<cstdio>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int MAXN = 21234567;
int euler[MAXN];
void get_euler()
{
_for(i, 1, MAXN) euler[i] = i;
_for(i, 2, MAXN)
if(euler[i] == i)
for(int j = i; j <= MAXN; j += i)
euler[j] = euler[j] / i * (i - 1);
}
void read(ll& x)
{
int f = 1; x = 0; char ch = getchar();
while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar(); }
while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
x *= f;
}
int main()
{
get_euler();
ll n, x; read(n);
while(n--)
{
read(x);
printf("%lld
", euler[x]);
}
return 0;
}
解法二
#include<cstdio>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
typedef long long ll;
const int MAXN = 21234567;
ll euler(ll x)
{
ll ret = x;
for(int i = 2; i * i <= x; i++)
if(x % i == 0)
{
ret = ret / i * (i - 1);
while(x % i == 0) x /= i;
if(x == 1) break;
}
if(x > 1) ret = ret / x * (x - 1);
return ret;
}
void read(ll& x)
{
int f = 1; x = 0; char ch = getchar();
while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar(); }
while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
x *= f;
}
int main()
{
ll n, x; read(n);
while(n--)
{
read(x);
printf("%lld
", euler(x));
}
return 0;
}