Duizi and Shunzi
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2015 Accepted Submission(s): 807
Problem Description
Nike likes playing cards and makes a problem of it.
Now give you n integers, ai(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, ai(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.
Input
The input contains several test cases.
For each test case, the first line contains one integer n( 1≤n≤106).
Then the next line contains n space-separated integers ai ( 1≤ai≤n)
For each test case, the first line contains one integer n( 1≤n≤106).
Then the next line contains n space-separated integers ai ( 1≤ai≤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
Recommend
题意：
给出n个数字，两个相同的数字可以组成对子，
三个连续的数字可以组成顺子，问能组成对子和顺子的总数最大是多少
思路：
统计每个元素的个数，从头开始循环，对于前两个元素，
判断能不能组成对子，能组成就加上相应的数量，并减去用掉的元素。
从第三个元素起，先检查第三个元素能不能与前面两个元素组成顺子，
然后再判断能不能组成对子，减去和加上相应的数量
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;//数字最大为10^6
int a[maxn];//a[i]表示i这个数字出现的次数
int main()
{
int n;//表示有n个数字
while(scanf("%d",&n)!=EOF){//多组样例测试模板
memset(a,0,sizeof(a));//初始化每个数字的个数为0
int duizi=0;//对子的个数
int shunzi=0;//顺子的个数
int x;//每个数字
for(int i=0;i<n;i++)//一共有n个数字
{
scanf("%d",&x);
a[x]++;//数字x的个数+1
}
for(int i=1;i<maxn;i++)//从前到后遍历每个容器
{
if(i<=2)//对于前两个元素
{
if(a[i]>=2){//判断能不能组成对子
int num=a[i]/2;//对子的个数
a[i]-=2*num;//并减去用掉的元素
duizi+=num;//能组成就加上相应的数量
}
}
else
{ //从第三个元素起，先检查第三个元素能不能与前面两个元素组成顺子
if(a[i]>0&&a[i-1]>0&&a[i-2]>0)
{
shunzi++;
a[i]--;a[i-1]--;a[i-2]--;
}
if(a[i]>=2){//判断能不能组成对子
int num=a[i]/2;
//减去和加上相应的数量
a[i]-=2*num;
duizi+=num;
}
}
}
printf("%d\n",duizi+shunzi);
}
return 0;
}