Duizi and Shunzi
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 661 Accepted Submission(s): 311
Problem Description
Nike likes playing cards and makes a problem of it.
Now give you n integers,
We define two identical numbers (eg: ) a Duizi,
and three consecutive positive integers (eg: ) a Shunzi.
Now you want to use these integers to form Shunzi and Duizi as many as possible.
Let s be the total number of the Shunzi and the Duizi you formed.
Try to calculate .
Each number can be used only once.
Now give you n integers,
We define two identical numbers (eg: ) a Duizi,
and three consecutive positive integers (eg: ) a Shunzi.
Now you want to use these integers to form Shunzi and Duizi as many as possible.
Let s be the total number of the Shunzi and the Duizi you formed.
Try to calculate .
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().
Then the next line contains n space-separated integers ()
For each test case, the first line contains one integer n().
Then the next line contains n space-separated integers ()
Output
For each test case, output the answer in a line.
Sample Input
71 2 3 4 5 6 7 91 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)
题意
给我们一串数字,让我们从这串数字中找出最多的对子和顺子
解题思路
排一下序,贪心的去利用这些数字,对子需要两个数,顺子需要三个数,所以应该优先贪心对子,这里有一个地方需要优先贪心顺子,就是当顺子已经集齐了两个,这个时候只需要一个数字就可以凑齐顺子,而对子需要两个数字才行
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std ; const int maxn = 1e6 + 10 ; int n , a[maxn] ; int main() { while(scanf("%d" , &n) != EOF) { for(int i = 0 ; i < n ; i++) scanf("%d" , a + i) ; sort(a , a + n) ; int stp = 0 ; int cnt = 0 ; int pre = -1 ; int sum = 0 ; while(stp < n) { // cout<< "stp : " << stp << "cnt : " << cnt << endl ; if(cnt == 2 && a[stp] == pre + 1)///顺子已经凑齐了两个差一个 { sum ++ ; cnt = 0 ; stp ++ ; continue ; } else if(stp == n - 1) break ; else if(a[stp] != a[stp + 1]) ///当前数字和下一个数字不同 { if(cnt == 0) ///如果顺子个数为0,当前数做顺子的第一个数 { cnt++ ; pre = a[stp] ; stp++ ; } else { if(a[stp] == pre + 1) ///如果当前数和顺子的前一个数能连续,顺子长度增加,更新顺子当前的数字 { cnt++ ; pre = a[stp] ; stp++ ; } else ///如果构不成连续,顺子长度从一重新开始 { cnt = 1 ; pre = a[stp] ; stp++ ; } } } else if(a[stp] == a[stp + 1])///贪心对子 { sum++ ; stp += 2 ; } } printf("%d\n" , sum) ; } return 0 ; }