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A9740. Candy Boxes

编程题 普及/提高-

题目描述

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 ${x_{1},x_{2},x_{3},x_{4}}$ ( $x_{1}<=x_{2}<=x_{3}<=x_{4}$ ) arithmetic mean is ![](/uploads/luogu/CF488B/4ced3a9ae4181924c136bcc46ede19844528fdef_c86ca214db32.png), median is ![](/uploads/acgo/image/39236d2f21ef7fbb_ba2cf5510dea.jpeg) and range is $x_{4}-x_{1}$ . 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)?

输入格式

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 ${x_{1},x_{2},x_{3},x_{4}}$ ( $x_{1}<=x_{2}<=x_{3}<=x_{4}$ ) arithmetic mean is ![](/uploads/luogu/CF488B/4ced3a9ae4181924c136bcc46ede19844528fdef_c86ca214db32.png), median is ![](/uploads/acgo/image/ef4aa68c2596f40a_248da7fd5b58.jpeg) and range is $x_{4}-x_{1}$ . 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)?

输出格式

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 ${x_{1},x_{2},x_{3},x_{4}}$ ( $x_{1}<=x_{2}<=x_{3}<=x_{4}$ ) arithmetic mean is ![](/uploads/luogu/CF488B/4ced3a9ae4181924c136bcc46ede19844528fdef_c86ca214db32.png), median is ![](/uploads/acgo/image/99dc730e2059639e_508bc7327854.jpeg) and range is $x_{4}-x_{1}$ . 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)?

输入输出样例

输入 #1
2
1
1
输出 #1
YES
3
3
输入 #2
3
1
1
1
输出 #2
NO
输入 #3
4
1
2
2
3
输出 #3
YES

说明/提示

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.
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