上午小测1「木板」
题目大意
分析
简单推论
老数学题了，直接开搞：
[设 ;BE = x,CE = n - x, CF = y, angle AEF = 90^{circ}
]
[ecause Delta ABE sim Delta ECF
]
[ herefore frac{AB}{BE} = frac{EC}{CF}
]
[ herefore frac{n}{x} = frac{n - x}{y}
]
[ herefore y = frac {nx-x^2}{n} = x - frac{x^2}{n}
]
[ecause xin mathbb{Z};且;yin mathbb{Z}
]
[ herefore n|x^2
]
[设;n=p_1^{a_1} imes p_2^{a_2} imes p_3^{a_3} imes cdots p_i^{a_i}(p_iin mathbb{P})
]
[ herefore x_{min} = p_1^{left lceil frac{a_1}{2}
ight
ceil} imes p_2^{left lceil frac{a_2}{2}
ight
ceil} imes p_3^{left lceil frac{a_3}{2}
ight
ceil} imes cdots imes p_i^{left lceil frac{a_i}{2}
ight
ceil}(p_iin mathbb{P})
]
[设; kin mathbb{Z}
]
[ecause kx_{min} < n
]
[ herefore k < frac{n}{x_{min}}
]
[易得; ans = k - 1
]
进一步计算
- 如何求 (x_{min}) ？
会发现，每个质因子次幂是向上取整的，当次幂是奇数的时候，有一个 (p_i) 是必须选取的，剩下的选一半即可。
[设; n=2^3 imes 3^2 imes 5^1
]
[x_{min} = 2^2 imes 3^1 imes 5^1
]
[有; 2^1; 和; 5^1; 是必选的
]
[设; i (i ^ 2leq n; 且; i^2|n), b = 2^1 imes 5^1
]
[ecause i_{max} = 2^1 imes 3^1
]
[ herefore b = frac{n}{i^2_{max}}
]
[ herefore x_{min} = i_{max}b = frac{n}{i_{max}}
]
[ecause k < frac{n}{x_{min}}
]
[ herefore k < i_{max}
]
[ecause ans = k - 1
]
[ herefore ans = i_{max} - 1
]
[ herefore ans = ans imes 8(正方形，四个角，直角在两边)
]
[证毕
]
代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;
const int maxn = 1e5 + 50, INF = 0x3f3f3f3f;
inline int read () {
register int x = 0, w = 1;
register char ch = getchar ();
for (; ch < '0' || ch > '9'; ch = getchar ()) if (ch == '-') w = -1;
for (; ch >= '0' && ch <= '9'; ch = getchar ()) x = x * 10 + ch - '0';
return x * w;
}
inline void write (register int x) {
if (x / 10) write (x / 10);
putchar (x % 10 + '0');
}
int n, ans;
signed main () {
freopen ("tri.in", "r", stdin);
freopen ("tri.out", "w", stdout);
while (1) {
n = read();
if (n == 0) break;
for (register int i = 1; i * i <= n; i ++) {
if (n % (i * i) == 0) ans = (i - 1) * 8;
}
printf ("%lld
", ans);
}
return 0;
}