uva 12493 - Red Huang

本來以為是一筆畫圖形，結果完全不是...

因為要走完所有的點而且每一步都是一樣長的，所以走的步數要跟總共的點互質

否則最大公因數如果不是 1 代表不到 n 步就走完了

其實是歐拉公式

取出 n 的所有質因數 a1,a2,a3

並且用公式 n2=n (1-1/a1)*(1-1/a2)*(1-1/a3)

這就等於把質因數全部的倍數刪掉取出剩下的數

//============================================================================  
// Name        : stars2.cpp  
// Date : 2013/3/22 上午11:47:39  
// Author : GCA  
//============================================================================  
#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <climits>  
#include <vector>  
#include <set>  
#include <map>  
#include <queue>  
#include <cctype>  
#include <utility>  
using namespace std;  
#ifdef ONLINE\_JUDGE  
#define ll "%lld"  
#else  
#define ll "%I64d"  
#endif  
typedef unsigned int uint;  
typedef long long int Int;  
#define Set(a,s) memset(a,s,sizeof(a))  
#define Write(w) freopen(w,"w",stdout)  
#define Read(r) freopen(r,"r",stdin)  
#define Pln() printf("\\n")  
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()  
#define De(x)printf(#x"%d\\n",x)  
#define For(i,x)for(int i=0;i<x;i++)  
#define CON(x,y) x##y  
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\  
    printf("%d ",dp\[htx\]\[hty\]);}Pln();}  
#define M 10000000  
#define PII pair<int,int\>  
#define PB push\_back  
#define oo INT\_MAX  
#define Set\_oo 0x3f  
#define Is\_debug true  
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)  
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)  
#define eps 1e-6  
bool xdy(double x,double y){return x>y+eps;}  
bool xddy(double x,double y){return x>y-eps;}  
bool xcy(double x,double y){return x<y-eps;}  
bool xcdy(double x,double y){return x<y+eps;}  
int min3(int x,int y,int z){  
    int tmp=min(x,y);  
    return min(tmp,z);  
}  
int max3(int x,int y,int z){  
    int tmp=max(x,y);  
    return max(tmp,z);  
}  
int n;  
vector<int\> a;  
void solve(){  
    int on=n;  
    int sqrtn=sqrt(n);  
    for(int i=2;i<=sqrtn;i++){  
        if(!(n%i)){  
            while(!(n%i)) n/=i;  
            a.PB(i);  
        }  
    }  
    if(n!=1)a.PB(n);  
    int size=a.size();  
    for(int i=0;i<size;i++){  
//        debug("%d\\n",a\[i\]);  
        on=on-on/a\[i\];  
    }  
    printf("%d\\n",on>>1);  
}  
int main() {  
    ios\_base::sync\_with\_stdio(0);  
    while(~scanf("%d",&n)){  
        solve();  
        a.clear();  
    }  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
}