Avito Cool Challenge 2018 E. Missing Numbers 【枚举】

传送门：http://codeforces.com/contest/1081/problem/E

E. Missing Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Chouti is working on a strange math problem.

There was a sequence of nn positive integers x1,x2,…,xnx1,x2,…,xn, where nn is even. The sequence was very special, namely for every integer ttfrom 11 to nn, x1+x2+...+xtx1+x2+...+xt is a square of some integer number (that is, a perfect square).

Somehow, the numbers with odd indexes turned to be missing, so he is only aware of numbers on even positions, i.e. x2,x4,x6,…,xnx2,x4,x6,…,xn. The task for him is to restore the original sequence. Again, it's your turn to help him.

The problem setter might make mistakes, so there can be no possible sequence at all. If there are several possible sequences, you can output any.

Input

The first line contains an even number nn (2≤n≤1052≤n≤105).

The second line contains n2n2 positive integers x2,x4,…,xnx2,x4,…,xn (1≤xi≤2⋅1051≤xi≤2⋅105).

Output

If there are no possible sequence, print "No".

Otherwise, print "Yes" and then nn positive integers x1,x2,…,xnx1,x2,…,xn (1≤xi≤10131≤xi≤1013), where x2,x4,…,xnx2,x4,…,xn should be same as in input data. If there are multiple answers, print any.

Note, that the limit for xixi is larger than for input data. It can be proved that in case there is an answer, there must be a possible sequence satisfying 1≤xi≤10131≤xi≤1013.

Examples

input

Copy

6
5 11 44

output

Copy

Yes
4 5 16 11 64 44

input

Copy

2
9900

output

Copy

Yes
100 9900

input

Copy

6
314 1592 6535

output

Copy

No

Note

In the first example

• x1=4x1=4
• x1+x2=9x1+x2=9
• x1+x2+x3=25x1+x2+x3=25
• x1+x2+x3+x4=36x1+x2+x3+x4=36
• x1+x2+x3+x4+x5=100x1+x2+x3+x4+x5=100
• x1+x2+x3+x4+x5+x6=144x1+x2+x3+x4+x5+x6=144

All these numbers are perfect squares.

In the second example, x1=100x1=100, x1+x2=10000x1+x2=10000. They are all perfect squares. There're other answers possible. For example, x1=22500x1=22500 is another answer.

In the third example, it is possible to show, that no such sequence exists.

题意概括：

给出 N/2 个 偶数位的数，需要补充剩余 N/2 个奇数位的数，使得 每一位的前缀和都是完全平方数。

如果存在解，按顺序输出 N 个完整数字。

解题思路：

从第一位开始构造前缀和，如果按照实际的数的范围去枚举 x（1e13）肯定会爆炸，那么就折半枚举√x，判断当前前缀和 Si 是否为完全平方数，

如果是则记录下来，并且从 √Si + 1开始枚举求解下一个前缀和。

如果超过的 √Smax ，说明无解。

AC code:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

const int MAXN = 1e5+10;
const LL MAXW = 1100000;

LL b[MAXN], a[MAXN];

bool check(LL x)
{
    LL y = sqrt(x);
    if(y*y == x) return true;
    return false;
}

int main()
{
    int N;
    LL sum = 0;
    scanf("%d", &N);
    for(int i = 1; i <= N/2; i++){
        scanf("%I64d", &a[i]);
    }

    LL tmp = 1LL;
    while(!check(tmp*tmp+a[1])){
        tmp++;
        if(tmp >= MAXW){
            puts("No");
            return 0;
        }
    }
    b[1] = tmp*tmp;
    b[2] = a[1];
    int cnt = 2;
    tmp = sqrt(tmp*tmp+a[1]);

    for(int i = 2; i <= N/2; i++){
        if(tmp >= MAXW){
                puts("No");
                return 0;
        }
        sum = tmp;
        tmp++;
        while(!check(tmp*tmp+a[i])){
            tmp++;
            if(tmp >= MAXW){
                puts("No");
                return 0;
            }
        }
        b[++cnt] = tmp*tmp - sum*sum;
        b[++cnt] = a[i];
        tmp = sqrt(tmp*tmp + a[i]);
        printf("a:%I64d tmp:%I64d
",a[i], tmp);
    }
    puts("Yes");
    for(int i = 1; i <= cnt; i++)
        printf("%I64d ", b[i]);

    return 0;
}