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[];
     int c[];
     cin >> n >> m;
     for (int i = ; i <= m; i++){
         c[i] = ;
     }
     for (int i = ; i < n; i++){
         cin >> a[i].color;
         a[i].num = i;
         c[a[i].color]++;
     }
     int mx = ;
     for (int i = ; i <= m; i++){
         if (c[mx] < c[i]){
             mx = i;
         }
     }
     if (c[mx] *  > n){
         cout <<  * (n - c[mx]) << endl;
         int mcount = , miter = -, d = n - c[mx];
         for (int i = ; i < n; i++){
             if (a[i].color == mx){
                 cout << a[i].color << " ";
                 if (d > ){
                     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 = ; i < n; i++){
             a[i].mat = a[(i + c[mx]) % n].color;
         }
         sort(a, a + n, sort_cmp);
         for (int i = ; i < n; i++){
             cout << a[i].color << " " << a[i].mat << endl;
         }
     }
     return ;
 }