2017ACM/ICPC广西邀请赛 1007 Duizi and Shunzi
Duizi and Shunzi
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total
Submission(s): 0 Accepted Submission(s): 0
Problem Description
Nike likes playing cards and makes a problem of
it.
Now give you n integers, a
i
(1≤i≤n)
We define two identical numbers (eg: 2,2
) a Duizi,
and three consecutive positive integers (eg: 2,3,4
) 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 max(s)
.
Each number can be used only once.
Now give you n integers, a
We define two identical numbers (eg: 2,2
and three consecutive positive integers (eg: 2,3,4
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 max(s)
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(1≤n≤10
6![]()
).
Then the next line contains n space-separated integers a
i![]()
(1≤a
i
≤n
)
For each test case, the first line contains one integer n(1≤n≤10
Then the next line contains n space-separated integers a
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)
题解:使用数组保存了每一个数字出现的次数
然后成1开始 循环到n 1ms的时间就可以了
dp记录的是前面循序的长度dp=0,1,2 ans记录的是答案 当你到了第i的数的时候 分以下情况
1》如果数据中没有i,也就是b[i]=0;那么把dp置0就好了
2》如果数据中有 然后dp=2了 那么不管有多少个i 我拿去一个i和前面的i-1,i-2放一起,形成一对 是最优的 (这你要明白)
3》如果数据中有 然后dp=0,1 要是i的数目是奇数 那就多出来的那个i 就和i-1放一起 dp++ 不然的话 dp=0 ans+=b[i]/2;
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <cstring> 6 #include <math.h> 7 using namespace std; 8 int b[1000010]; 9 //int qyh[10010],hyh[10010]; 10 int main() 11 { 12 int a,n; 13 while(~scanf("%d",&n)) 14 { 15 for(int i=0; i<1000010; ++i) 16 { 17 b[i]=0; 18 } 19 for(int i=0; i<n; ++i) 20 { 21 scanf("%d",&a); 22 b[a]++; 23 } 24 int dp=0; 25 int ans=0; 26 for(int i=1; i<=n; ++i) 27 { 28 if(b[i]==0) 29 { 30 dp=0; 31 continue; 32 } 33 if(dp==2) 34 { 35 ans+=1; 36 ans+=(b[i]-1)/2; 37 dp=(b[i]-1)%2; 38 } 39 else if(dp==1) 40 { 41 if(b[i]%2==1) 42 { 43 dp++; 44 ans+=b[i]/2; 45 } 46 else 47 { 48 dp=0; 49 ans+=b[i]/2; 50 } 51 } 52 else if(dp==0) 53 { 54 if(b[i]%2==1) 55 { 56 dp++; 57 ans+=b[i]/2; 58 } 59 else 60 { 61 dp=0; 62 ans+=b[i]/2; 63 } 64 } 65 } 66 printf("%d\n",ans); 67 } 68 return 0; 69 }
作者:红雨
出处:https://www.cnblogs.com/52why
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