Reflect

Reflect

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 288    Accepted Submission(s): 174

Problem Description

We send a light from one point on a mirror material circle,it reflects N times and return the original point firstly.Your task is calcuate the number of schemes.

 

Input

First line contains a single integer T(T≤10) which denotes the number of test cases.

For each test case, there is an positive integer N(N≤106).

 

Output

For each case, output the answer.

 

Sample Input

14

 

Sample Output

4

 





白痴互质求法：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <queue>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define eps 1e-8
#define INF 20000000005
#define rd(x) scanf("%d",&x)
#define rdLL(x) scanf("%I64d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long
#define mod 998244353
#define maxn 100005
#define maxm 1000
#define minn 0.00000001;

bool judge(int a,int b){
   int temp=0;
    while(b!=0){
        temp=b;
        b=a%b;
        a=temp;
    }
    if(a==1) return 1;//如果最大公约数是1，那么两数互质
    else return 0;
}

int main()
{
    int Case,n;
    rd(Case);
    while(Case--){
        rd(n);
        int coun = 1;
        for(int i=2 ; i<n+1 ; i++)
            if( judge(i,n+1) )
                coun++;
        printf("%d
",coun);
    }
    return 0;
}

欧拉函数：（两种方式）

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <queue>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define eps 1e-8
#define INF 20000000005
#define rd(x) scanf("%d",&x)
#define rdLL(x) scanf("%I64d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long
#define mod 998244353
#define MAX 1000005
#define maxm 1000
#define minn 0.00000001;

///但个数的的欧拉函数
//long long eular(long long n){
//  long long ans = n;
//  for(int i=2 ; i*i <= n ; i++){
//     if(n%i == 0){
//        ans = ans - ans/i;  ///欧拉函数
//        while(n%i == 0)
//            n = n/i;
//     }
//  }
//  if( n>1 ) ans=ans-ans/n;
//  return ans;
//}

///筛法欧拉函数（打表）
int eular [MAX];
void getEular(){
   memset(eular,0,sizeof(eular));
   eular[1] = 1;
   for(int i=2;i<=MAX;i++){
    if(!eular[i])
      for(int j=i;j<=MAX ; j += i){
        if(!eular[j]) eular[j] = j;
        eular[j] = eular[j]/i*(i-1);
      }
   }
}

int main()
{
    int Case,n;
    getEular();
    rd(Case);
    while(Case--){
        rd(n);
        printf("%d
",eular[n+1]);
        //printf("%lld
",eular(n+1));
    }
    return 0;
}