CodeForces 370C

这道题其实是挺简单的。首先很容易发现最多人用的颜色的人数如果大于n/2，就肯定不能让全部人都成功戴上两只不同颜色的手套。反过来想，如果这个人数小于等于n/2又如何呢？的确，这就能让全部人都能戴上两只不同颜色的手套（每个人都留一只自己原本的，再要别人的一只手套就可以了）。

#include <iostream>
#include <cmath>
#include <algorithm>
#include <bitset>
#include <cstring>
#include <list>
using namespace std;
struct Child{
    int color, num, mat;
};
bool partition_cmp(const Child& c1, const Child& c2){
    return c1.color < c2.color;
}
bool sort_cmp(const Child& c1, const Child& c2){
    return c1.num < c2.num;
}
int main(){
    int n, m;
    Child a[5000];
    int c[101];
    cin >> n >> m;
    for (int i = 1; i <= m; i++){
        c[i] = 0;
    }
    for (int i = 0; i < n; i++){
        cin >> a[i].color;
        a[i].num = i;
        c[a[i].color]++;
    }
    int mx = 0;
    for (int i = 1; i <= m; i++){
        if (c[mx] < c[i]){
            mx = i;
        }
    }
    if (c[mx] * 2 > n){
        cout << 2 * (n - c[mx]) << endl;
        int mcount = 0, miter = -1, d = n - c[mx];
        for (int i = 0; i < n; i++){
            if (a[i].color == mx){
                cout << a[i].color << " ";
                if (d > 0){
                    for (miter++; miter < n && a[miter].color == mx; miter++);
                    cout << a[miter].color << endl;
                    d--;
                }
                else{
                    cout << a[i].color << endl;
                }
            }
            else{
                cout << a[i].color << " " << mx << endl;
            }
        }
    }
    else{
        cout << n << endl;
        sort(a, a + n, partition_cmp);
        for (int i = 0; i < n; i++){
            a[i].mat = a[(i + c[mx]) % n].color;
        }
        sort(a, a + n, sort_cmp);
        for (int i = 0; i < n; i++){
            cout << a[i].color << " " << a[i].mat << endl;
        }
    }
    return 0;
}