Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 153 Accepted Submission(s): 71
Problem Description
Nike likes playing cards and makes a problem of it.
Now give you n integers, ).
Each number can be used only once.
Now give you n integers, ).
Each number can be used only once.
Input
The input contains several test cases.
For each test case, the first line contains one integer n(n)
For each test case, the first line contains one integer n(n)
Output
For each test case, output the answer in a line.
Sample Input
7
1 2 3 4 5 6 7
9
1 1 1 2 2 2 3 3 3
6
2 2 3 3 3 3
6
1 2 3 3 4 5
Sample Output
2
4
3
2
Hint
Case 1（1，2，3）（4，5，6） Case 2（1，2，3）（1，1）（2，2）（3，3） Case 3（2，2）（3，3）（3，3） Case 4（1，2，3）（3，4，5）
Source
1 //2017-08-31 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 7 using namespace std; 8 9 const int N = 1100000; 10 int arr[N], n; 11 bool book[N]; 12 13 int main() 14 { 15 //freopen("input1007.txt", "r", stdin); 16 while(scanf("%d", &n) != EOF){ 17 for(int i = 0; i < n; i++) 18 scanf("%d", &arr[i]); 19 sort(arr, arr+n); 20 memset(book, 0, sizeof(book)); 21 int ans = 0; 22 for(int i = 1; i < n; i++){ 23 if(i >= 2){ 24 int p1 = -1, p2 = -1; 25 for(int j = i-1; j >= 0; j--){ 26 if(arr[j] == arr[i]-1 && !book[j]){ 27 p1 = j; 28 } 29 if(arr[j] == arr[i]-2 && !book[j]){ 30 p2 = j; 31 break; 32 } 33 if(arr[j] < arr[i]-1)break; 34 } 35 if(p1 != -1 && p2 != -1){ 36 ans++; 37 book[i] = book[p1] = book[p2] = 1; 38 } 39 } 40 if(arr[i-1] == arr[i] && !book[i-1] && !book[i]){ 41 ans++; 42 book[i-1] = book[i] = 1; 43 } 44 } 45 printf("%d\n", ans); 46 } 47 48 return 0; 49 }