Codeforces Round #278 (Div. 2) B. Candy Boxes [brute force+constructive algorithms]

哎，最近弱爆了，，，不过这题还是不错滴~~ 要考虑完整各种情况

B. Candy Boxes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {x₁, x₂, x₃, x₄} (x₁ ≤ x₂ ≤ x₃ ≤ x₄) arithmetic mean is , median is and range is x₄ - x₁. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.

For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.

Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n (0 ≤ n ≤ 4) boxes remaining. The i-th remaining box contains a_i candies.

Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?

Input

The first line of input contains an only integer n (0 ≤ n ≤ 4).

The next n lines contain integers a_i, denoting the number of candies in the i-th box (1 ≤ a_i ≤ 500).

Output

In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.

If a solution exists, you should output 4 - n more lines, each line containing an integer b, denoting the number of candies in a missing box.

All your numbers b must satisfy inequality 1 ≤ b ≤ 10⁶. It is guaranteed that if there exists a positive integer solution, you can always find such b‘s meeting the condition. If there are multiple answers, you are allowed to print any of them.

Given numbers a_i may follow in any order in the input, not necessary in non-decreasing.

a_i may have stood at any positions in the original set, not necessary on lowest n first positions.

Sample test(s)

Input

2 1 1

Output

YES 3 3

Input

3 1 1 1

Output

NO

Input

4 1 2 2 3

Output

YES

Note

For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.

For the second sample, it‘s impossible to find the missing number of candies.

In the third example no box has been lost and numbers satisfy the condition.

You may output b in any order.

主要就是根据那三个等式，化简，得两个方程：

x4=3*x1;

x2+x3=4*x1;

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<string>
 11 //#include<pair>
 12
 13 #define N 10005
 14 #define M 1005
 15 #define mod 1000000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define LL long long
 20 #define eps 1e-9
 21 #define maxi(a,b) (a)>(b)? (a) : (b)
 22 #define mini(a,b) (a)<(b)? (a) : (b)
 23
 24 using namespace std;
 25
 26 int n;
 27 int a[10];
 28 int b[10];
 29 int flag;
 30
 31 void ini()
 32 {
 33     flag=0;
 34     for(int i=1;i<=n;i++){
 35         scanf("%d",&a[i]);
 36     }
 37     sort(a+1,a+1+n);
 38 }
 39
 40 void solve0()
 41 {
 42     flag=1;
 43     b[1]=1;b[2]=1;b[3]=3;b[4]=3;
 44 }
 45
 46 void solve1()
 47 {
 48     flag=1;
 49     b[1]=a[1];b[2]=3*a[1];b[3]=3*a[1];
 50 }
 51
 52 void solve2()
 53 {
 54     if(a[2]%3==0){
 55         if(a[2]/3==a[1]){
 56             flag=1;
 57             b[1]=a[1];b[2]=a[2];
 58         }
 59         else if(a[2]/3<a[1]){
 60             flag=1;b[1]=a[2]/3;b[2]=a[2]+b[1]-a[1];
 61         }
 62         else{
 63             return;
 64         }
 65     }
 66     else{
 67         if(a[1]*3<=a[2]){
 68             return;
 69         }
 70         else{
 71             flag=1;
 72             b[2]=a[1]*3;
 73             b[1]=b[2]+a[1]-a[2];
 74         }
 75     }
 76 }
 77
 78 void solve3()
 79 {
 80     if(a[3]%3!=0){
 81         if(a[1]*3<=a[3]){
 82             return;
 83         }
 84         else{
 85             if(a[1]*4==a[2]+a[3]){
 86                 flag=1;b[1]=a[1]*3;
 87             }
 88             else return;
 89         }
 90     }
 91     else{
 92         if(a[3]/3==a[1]){
 93             flag=1;
 94             b[1]=a[1]+a[3]-a[2];
 95         }
 96         else if(a[3]/3<a[1]){
 97             b[1]=a[3]/3;
 98             if(b[1]+a[3]==a[1]+a[2]){
 99                 flag=1;
100             }
101             else return;
102         }
103         else{
104             return;
105         }
106     }
107 }
108
109 void solve4()
110 {
111     if(a[4]==3*a[1] && a[2]+a[3]==4*a[1]){
112         flag=1;
113     }
114 }
115
116 void solve()
117 {
118     if(n==4){
119         solve4();
120     }
121     else if(n==0){
122         solve0();
123     }
124     else if(n==1){
125         solve1();
126     }
127     else if(n==2){
128         solve2();
129     }
130     else if(n==3){
131         solve3();
132     }
133 }
134
135 void out()
136 {
137     int i;
138     if(flag==1){
139         printf("YES\n");
140         for(i=1;i<=4-n;i++){
141             printf("%d\n",b[i]);
142         }
143     }
144     else{
145         printf("NO\n");
146     }
147 }
148
149 int main()
150 {
151     //freopen("data.in","r",stdin);
152     //freopen("data.out","w",stdout);
153     //scanf("%d",&T);
154     //for(int ccnt=1;ccnt<=T;ccnt++)
155    // while(T--)
156     while(scanf("%d",&n)!=EOF)
157     {
158        // if(n==0 && m==0 ) break;
159         //printf("Case %d: ",ccnt);
160         ini();
161         solve();
162         out();
163     }
164
165     return 0;
166 }