一道难题？主要是木有方法。
1041. Be Unique (20)
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” instead.
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
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！
You deserve a spot in the OP too!
)