LuoGu: CF1081E Missing Numbers
CF: E. Missing Numbers
考虑前面 $i - 1$ 个数都是合法的，现在加入第 $i$ 个数 $w$ 。
那么有 $a ^ 2 - b ^ 2 = w$ ，即 $(a + b) (a - b) = w$ 。将 $w$ 分解质因数即可解得两个答案，注意保证答案合法。因为对答案有大小限制，所有尽量保证答案尽可能小，所以分解质因数时优先考虑两个因数差更小的情况，这样平方会更小。
查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
| #include <cstdio>
#include <cmath>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e5 + 10;
LL w[N], v[N];
int n;
int main ()
{
read(n);
for (int i = 2; i <= n; i += 2)
{
read(w[i]);
bool flag = false;
for (int j = sqrt(w[i]) + 0.5; j && !flag; j--)
{
if (w[i] % j || (j & 1) ^ (w[i] / j & 1))
continue;
v[i] = w[i] / j + j >> 1;
v[i - 1] = w[i] / j - j >> 1;
flag = v[i - 1] > v[i - 2] && v[i] * v[i] > w[i];
}
if (!flag)
return puts("No"), 0;
}
for (int i = 1; i <= n; i += 2)
w[i] = v[i] * v[i] - v[i - 1] * v[i - 1];
puts("Yes");
for (int i = 1; i <= n; i++)
write(w[i]), putchar(' ');
return 0;
}
|