一道难题？主要是木有方法。

1041. Be Unique (20)

时间限制

20 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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, 10⁴]. 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 (<=10⁵) 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” instead.
Sample Input 1:

7 5 31 5 88 67 88 17

Sample Output 1:

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None

本人解答：

//#include 
//#include 
#include 
#include 
using namespace std;
const int MAX_NUM_A = 10001;
//const int MAX_NUM_B = 10000;
int storage[MAX_NUM_A] = {0};
//list possibles;
int possibles[MAX_NUM_A] = {0};
//0代表尚未存储，-1代表已经超额，其余代表输入时的顺序（第几个输入的这个数）
//int sequence[MAX_NUM_B];
//list ary_list;
/*
inline void remove(int r)
{
	//remove item from array
	ary_list.remove(r);
}
*/
int main()
{
	clock_t  clockBegin, clockMid, clockEnd;
	int possibleN = 0;
	int N;
	//cin >> N;
	scanf("%u",&N);
	int input;
	clockBegin = clock();
	for (int i=0; i> input;
		scanf("%u", &input);
		if(storage[input] == 0)
		{
			//尚未输入过这个数
			storage[input] = i+1; //记录输入的顺序
			//possibleN ++;
			//possibles.push_back(input);
			possibles[possibleN++] = input;
		}
		else
		{
			//重复了！
			//if(storage[input] > 0)
			//{
				//possibles.remove(input);
				//possibleN --;
			//}
			storage[input] = -1;
		}
	}
	/*
	if(possibles.size() <= 0)
	{
		cout << "None";
		return 0;
	}
	*/
	clockMid = clock();
	//cout << clockMid - clockBegin << endl;
	printf("%un",clockMid - clockBegin);
	//int count = ary_list.size();
	int min_seq = MAX_NUM_A;
	int min_val = -1;
	int i=0;
	while(i < possibleN)
	{
		int temp = storage[possibles[i]];
		if(temp > 0)
		{
			//if(temp < min_seq)
			//{
				min_seq = temp;
				min_val = possibles[i];
				break;
			//}
		}
		//possibles.pop_front();
		i ++ ;
	}
	clockEnd = clock();
	//cout << clockEnd - clockMid << endl;
	printf("%un",clockEnd - clockMid);
	if(min_val != -1)
	{
		//cout << min_val;
		printf("%u",min_val);
		return 0;
	}
	//cout << "None";
	printf("None");
	return 0;
}

知道结论是什么么？
结论是sscan比cin效率高很多！

1041. Be Unique (20)

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