2017广西邀请赛/hdu

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Duizi and Shunzi

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 245    Accepted Submission(s): 115


Problem Description
Nike likes playing cards and makes a problem of it.

Now give you n integers, ai(1in)

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(1n106). 
Then the next line contains n space-separated integers ai (1ain)
 

Output
For each test case, output the answer in a line.
 

Sample Input
71 2 3 4 5 6 791 1 1 2 2 2 3 3 362 2 3 3 3 3 61 2 3 3 4 5
 

Sample Output
2432
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
2017ACM/ICPC广西邀请赛-重现赛(感谢广西大学)
 

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Statistic | Submit | Discuss | Note
输入一个n,接下来有n个数,让你求出能组成最多的对子或者顺子的和。 
对子: (2,2),顺子: (1,2,3)。

贪心思路,优先考虑对子,其次是顺子,详见代码

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)#define eps 1e-8//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC<<endl;typedef long long ll;typedef long long LL;using namespace std;typedef unsigned long long ull;const int maxn=1e6+10;int a[maxn];int main(){    ios::sync_with_stdio(false);    int n;    while(cin>>n)    {        int x;        memset(a,0,sizeof(a));        for(int i=0;i<n;i++)            cin>>x,a[x]++;        int ans=0;        for(int i=1;i<=n;i++)        {            ans+=(a[i]/2);            a[i]%=2;            if(i+1<n)            {                if(a[i]==1 && a[i+1]%2 && a[i+2])                {                    ans++;                    a[i]--;                    a[i+1]--;                    a[i+2]--;                }            }        }        cout<<ans<<endl;    }    return 0;}

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