CodeForces 370C

這道題實際上是挺簡單的。首先很容易發現最多人用的顏色的人數若是大於n/2，就確定不能讓所有人都成功戴上兩隻不一樣顏色的手套。反過來想，若是這我的數小於等於n/2又如何呢？的確，這就能讓所有人都能戴上兩隻不一樣顏色的手套（每一個人都留一隻本身本來的，再要別人的一隻手套就能夠了）。

 1 #include <iostream>
 2 #include <cmath>
 3 #include <algorithm>
 4 #include <bitset>
 5 #include <cstring>
 6 #include <list>
 7 using namespace std;
 8 struct Child{
 9     int color, num, mat;
10 };
11 bool partition_cmp(const Child& c1, const Child& c2){
12     return c1.color < c2.color;
13 }
14 bool sort_cmp(const Child& c1, const Child& c2){
15     return c1.num < c2.num;
16 }
17 int main(){
18     int n, m;
19     Child a[5000];
20     int c[101];
21     cin >> n >> m;
22     for (int i = 1; i <= m; i++){
23         c[i] = 0;
24     }
25     for (int i = 0; i < n; i++){
26         cin >> a[i].color;
27         a[i].num = i;
28         c[a[i].color]++;
29     }
30     int mx = 0;
31     for (int i = 1; i <= m; i++){
32         if (c[mx] < c[i]){
33             mx = i;
34         }
35     }
36     if (c[mx] * 2 > n){
37         cout << 2 * (n - c[mx]) << endl;
38         int mcount = 0, miter = -1, d = n - c[mx];
39         for (int i = 0; i < n; i++){
40             if (a[i].color == mx){
41                 cout << a[i].color << " ";
42                 if (d > 0){
43                     for (miter++; miter < n && a[miter].color == mx; miter++);
44                     cout << a[miter].color << endl;
45                     d--;
46                 }
47                 else{
48                     cout << a[i].color << endl;
49                 }
50             }
51             else{
52                 cout << a[i].color << " " << mx << endl;
53             }
54         }
55     }
56     else{
57         cout << n << endl;
58         sort(a, a + n, partition_cmp);
59         for (int i = 0; i < n; i++){
60             a[i].mat = a[(i + c[mx]) % n].color;
61         }
62         sort(a, a + n, sort_cmp);
63         for (int i = 0; i < n; i++){
64             cout << a[i].color << " " << a[i].mat << endl;
65         }
66     }
67     return 0;
68 }