LA UVaLive 7372 Excellence (水题，贪心)

题意：给定 n 个数，要求把其中两个分成一组，然后加和，问所有的都分好，最小数是几。

析：贪心策略，最大和是小的相加，就是最优的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <functional>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <deque>
#include <map>
#include <cctype>
#include <stack>
#include <sstream>
using namespace std ;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
int a[maxn];

int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 0; i < n; ++i)  scanf("%d", &a[i]);
        sort(a, a+n);
        int ans = INF;
        for(int i = 0; i < n/2; ++i){
            ans = Min(ans, a[i]+a[n-i-1]);
        }
        printf("%d\n", ans);
    }
    return 0;
}