http://codeforces.com/contest/621/problem/A
A. Wet Shark and Odd and Even
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.
Input
The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Sample test(s)
input
3
1 2 3
output
6
input
5
999999999 999999999 999999999 999999999 999999999
output
3999999996
Note
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
题意,n个数，每个数最多使用一次，问能得到的最大的偶数和是多少。 思路：偶数不影响。偶数个奇数和还是偶数，不影响。奇数个奇数的时候把最小的奇数拿出来。
1/* ***********************************************
2Author :111qqz
3Created Time :2016年01月31日 星期日 22时02分34秒
4File Name :code/cf/#341/A.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E5+7;
34
35LL n;
36LL a[N];
37int main()
38{
39 #ifndef ONLINE_JUDGE
40 freopen("code/in.txt","r",stdin);
41 #endif
42 cin>>n;
43 int cnt = 0 ;
44 LL minodd = inf;
45 LL ans = 0 ;
46 for ( int i = 0 ; i < n ; i++)
47 {
48 scanf("%lld",&a[i]);
49 if (a[i]%2==1)
50 {
51 cnt++;
52 minodd = min(minodd,a[i]);
53 }
54 ans +=a[i];
55 }
56 if (cnt%2==1) ans-=minodd;
57 cout<<ans<<endl;
58
59 #ifndef ONLINE_JUDGE
60 fclose(stdin);
61 #endif
62 return 0;
63}