Day1-D-CF-1144C

简述：给你一个数组，判断是否能拆分成2个数组，一个递增一个递减，若不行输出No，可以就Yes并分别输出

思路：统计每个数出现的次数，若有大于2的肯定无法组成严格单调，这样就只需要将出现两次的组成递，剩下的递减即可

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<set>
#include<map>

using namespace std;

//const int maxm = 2019;
//int buff[maxm];
map<int,int> sum;

int main() {
    int n;
    scanf("%d",&n);
    getchar();
    for(int i = 0;i < n;++i) {
        int tmp;
        scanf("%d",&tmp);
        ++sum[tmp];
    }
    bool flag = true;
    int dbnum = 0;
    for(auto i = sum.begin();i != sum.end(); ++i) {
        if(i->second == 2)++dbnum;
        if(i->second > 2) {
            flag = false;
            break;
        }
    }
    if(!flag)
        printf("NO");
    else {
        printf("YES
");
        printf("%d
",dbnum);
        for(auto i = sum.begin(); i != sum.end(); ++i) {
            if(i->second == 2)
                cout << i->first << ' ';
        }
        printf("
");
        printf("%d
",n-dbnum);
        for(auto i = sum.rbegin();i != sum.rend();++i) {
            cout << i->first << ' ';
        }
        printf("
");
    }
    return 0;
}