solution
已知 a 1 , a 2 , a 3 , a 4 , . . . , a n − 1 , a n , 前 缀 和 s u m 2 [ i ] 均 为 平 方 数 ， a_1,a_2,a_3,a_4,...,a_{n-1},a_n,前缀和sum^2[i]均为平方数， a1,a2,a3,a4,...,an−1,an,前缀和sum2[i]均为平方数，
{ s u m [ 2 k + 1 ] = s u m 2 [ 2 k ] + x s u m [ 2 k + 2 ] = s u m 2 [ 2 k + 1 ] + y \begin{cases}sum[2k+1]=\sqrt{sum^2[2k] + x} \\sum[2k+2]=\sqrt{sum^2[2k+1] + y} \end{cases} { sum[2k+1]=sum2[2k]+xsum[2k+2]=sum2[2k+1]+y
= > => =>
{ s u m [ 2 k + 2 ] = s u m 2 [ 2 k + 1 ] + x + y s u m 2 [ 2 k + 2 ] ≥ ( s u m [ 2 k + 1 ] + 1 ) 2 y ≥ 2 ∗ s u m [ 2 k + 1 ] + 1 \begin{cases} sum[2k+2]=\sqrt{sum^2[2k+1] +x+ y} \\sum^2[2k+2]≥(sum[2k+1]+1)^2 \\y≥2*sum[2k+1]+1 \end{cases} ⎩⎪⎨⎪⎧sum[2k+2]=sum2[2k+1]+x+ysum2[2k+2]≥(sum[2k+1]+1)2y≥2∗sum[2k+1]+1
= > => =>
a [ 2 k + 2 ] ≥ 2 ∗ s u m [ 2 k + 1 ] + 1 a[2k+2]≥2*sum[2k+1]+1 a[2k+2]≥2∗sum[2k+1]+1
code
/*Siberian Squirrel*/
/*Cute JinFish*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const double PI = acos(-1), eps = 1e-8;
/*const int MOD = 998244353, r = 119, k = 23, g = 3;
const int MOD = 1004535809, r = 479, k = 21, g = 3;*/
const int MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const int N = 2e6 + 10, M = 1e7 + 10;
int sgn(double x) {
if(fabs(x) < eps) return 0;
return x < 0? -1: 1;
}
//inline int rnd(){static int seed=2333;return seed=(((seed*666666ll+20050818)%998244353)^1000000007)%1004535809;}
//double Rand() {return (double)rand() / RAND_MAX;}
int n;
ll a[N];
void init() {
}
void solve(ll res = 0, ll st = 0, bool f = true) {
for(int i = 1; i <= n; i += 2) {
for(ll j = st + 1; ; ++ j) {
ll x = j * j - st * st, y = a[i + 1];
ll now = sqrt(j * j + a[i + 1]);
if(now * now == j * j + a[i + 1]) {
a[i] = x;
st = now;
break;
}
if(y < 2 * now + 1) {
f = false;
break;
}
}
}
if(!f) cout << "No" << endl;
else {
cout << "Yes" << endl;
for(int i = 1; i <= n; ++ i) {
if(i == 1) cout << a[i];
else cout << ' ' << a[i];
}
cout << endl;
}
}
int main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(nullptr);
// srand(time(0));
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
init();
int o = 1;
// cin >> o;
while(o --) {
cin >> n;
for(int i = 1; i <= n / 2; ++ i) cin >> a[i << 1];
solve();
}
return 0;
}