1041 Be Unique （20 分）

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:

Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10​5​​) and then followed by N bets. The numbers are separated by a space.

Output Specification:

For each test case, print the winning number in a line. If there is no winner, print None

Sample Input 1:

7 5 31 5 88 67 88 17

Sample Output 1:

31

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None

題目大意：輸出第一個，在序列中沒有重複出現的數。

解題思路：以前有一道比較難的，輸出一個在序列中出現次數為奇數的，用異或就可以解決了，這裡的比較簡單，簡單用一下map對映到這個數出現的次數就可以了。因為時間限制只能進行一次迴圈。

程式碼如下：

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
map<int, int> a;
int num[100005];
int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> num[i];
		a[num[i]] += 1;
	}
	bool flag = false;
	for (int i = 0; i < n; i++) {
		if (a[num[i]] == 1) {
			flag = true;
			cout << num[i] << endl;
			break;
		}
	}
	if (!flag)cout << "None" << endl;
	return 0;
}
 

1042 Shuffling Machine （20 分）

Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.

The machine shuffles a deck of 54 cards according to a given random order and repeats for a given number of times. It is assumed that the initial status of a card deck is in the following order:

S1, S2, ..., S13, 
H1, H2, ..., H13, 
C1, C2, ..., C13, 
D1, D2, ..., D13, 
J1, J2

where "S" stands for "Spade", "H" for "Heart", "C" for "Club", "D" for "Diamond", and "J" for "Joker". A given order is a permutation of distinct integers in [1, 54]. If the number at the i-th position is j, it means to move the card from position i to position j. For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer K (≤20) which is the number of repeat times. Then the next line contains the given order. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the shuffling results in one line. All the cards are separated by a space, and there must be no extra space at the end of the line.

Sample Input:

2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47

Sample Output:

S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5

題目大意：模擬一下洗牌過程，英文是看不太懂的一開始，後來看到這54張牌大概就懂了。

解題思路：暴力過，每次洗牌的過程都用個臨時的vector記錄還沒開始洗的上一次結果，然後直接把上一次結果的陣列根據你輸入的洗牌順序修改，這樣即使是20次，規模也就是在1000左右，400ms的時間限制肯定是綽綽有餘的。

程式碼如下：

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
vector<int> order;
vector<string> s;
void init() {
	for (int i = 1; i <= 13; i++) {
		s.push_back("S" + to_string(i));
	}
	for (int i = 1; i <= 13; i++) {
		s.push_back("H" + to_string(i));
	}
	for (int i = 1; i <= 13; i++) {
		s.push_back("C" + to_string(i));
	}
	for (int i = 1; i <= 13; i++) {
		s.push_back("D" + to_string(i));
	}
	s.push_back("J1");
	s.push_back("J2");
}
int main()
{
	init();
	int K;
	cin >> K;
	for (int i = 0; i < 54; i++) {
		int c;
		cin >> c;
		order.push_back(c);
	}
	for (int i = 0; i < K; i++) {
		vector<string> tmp = s;
		for(int j=0;j<54;j++)
		s[order[j]-1] = tmp[j];
	}
	for (int i = 0; i < 54; i++) {
		if (i != 53)
			cout << s[i] << " ";
		else
			cout << s[i] << endl;
	}
	return 0;
}