Keep On Movin

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5744

Description

Professor Zhang has kinds of characters and the quantity of the i-th character is ai. Professor Zhang wants to use all the characters build several palindromic strings. He also wants to maximize the length of the shortest palindromic string.

For example, there are 4 kinds of characters denoted as ‘a’, ‘b’, ‘c’, ‘d’ and the quantity of each character is {2,3,2,2} . Professor Zhang can build {“acdbbbdca”}, {“abbba”, “cddc”}, {“aca”, “bbb”, “dcd”}, or {“acdbdca”, “bb”}. The first is the optimal solution where the length of the shortest palindromic string is 9.

Note that a string is called palindromic if it can be read the same way in either direction.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤105) — the number of kinds of characters. The second line contains n integers a1,a2,…,an (0≤ai≤104).

Output

For each test case, output an integer denoting the answer.

Sample Input

4
4
1 1 2 4
3
2 2 2
5
1 1 1 1 1
5
1 1 2 2 3

Sample Output

3
6
1
3

Hint

题意

有n个字符，每个字符ai个，你需要构造一些字符串，使得这些字符串都是回文串，而且这些回文串恰好用完所有字符

而且最短的回文串最长，输出这个长度。

题解：

贪心去构造，考虑奇数的字符，我们可以拆成偶数+1，那么显然我们知道，奇数的就一定是单独的一行，然后偶数一定要成对的扔进去。

然后考虑到这些公式就很简单了……

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int a[maxn];
void solve(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    int num = 0;
    long long ans = 0;
    long long sum = 0;
    long long Ans = 0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]%2==1)num++;
        ans+=a[i];
        Ans+=a[i]/2;
    }
    if(num==0){
        cout<<ans<<endl;
    }else{
        cout<<Ans/num*2+1<<endl;
    }
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--)solve();
}