zoj3590 -3+1

-3+1

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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ZOJ is 10 years old! For celebrating, we are offering the easiest problem in the world to you.

Recently we received a long sequence. We can modify the sequence once by the following two steps.

1. Choose any element in the sequence, say x(satisfying x ≥ 3), and subtract 3 from x.
2. Choose any element in the sequence, say x, and add 1 to x.

 

Now, we want to know how many times at most the sequence can be modified.

Input

The input contains multiple test cases. For each case, the first line contains an integer n(1 ≤ n ≤ 20000). The second line contains n integers describing the sequence. All the numbers in the sequence are non-negative and not greater than 1000000.

Output

Output number of times at most the sequence can be modified, one line per case.

Sample Input

1
10
2
10 11

Sample Output

4
10
题意：给你n个数，每次操作包括一次对其中一个数进行 -3 和  一个数进行 +1 ， 最多可以进行多少次操作；

分析：nn 代表 n 个数每个除 3 之和，n1 表示这 n 个数中模 3 余 1 的个数，n2 表示 mod 3 == 1 的个数；

#include<iostream>
using namespace std;
int main()
{
    long long n,n1,n2,nn,i,ans,t;
    while(cin>>n)
    {
        ans=0;n1=n2=nn=0;
        for(i=0;i<n;i++)
        {
            cin>>t;
            ans=ans+t/3;
            if(t%3==1) n1++;
            if(t%3==2) n2++;
        }
        nn=ans;
        if(nn<1)
        {
            cout<<ans<<endl;
        }
        else
        {
            ans=ans+n2;
            if(nn<=n1)
            {
                ans=ans+(nn-1);
                cout<<ans<<endl;
            }
            else
            {
                ans=ans+n1;
                nn=nn-n1;
                ans=ans+(nn-1)/2;
                cout<<ans<<endl;
            }
        }
    }
    return 0;
}