HDU 6222 Heron and His Triangle (pell 方程)

题面（本人翻译）

A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t - 1, t, t + 1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger

than or equal to n.

一个三角形是 Heron 三角形仅当它的三边长是连续的正整数 t - 1, t, t + 1， 并且面积是正整数。现在，给你一个整数 N ，求大于等于 N 的最小的合法的 t （Heron 三角形的第二小的边）。

Input

The input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30).

一个正整数 T 表示数据组数，1 ≤ T ≤ 30000。接下来 T 行每行一个整数 N，1 ≤ N ≤ 10^30。

Output

For each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.

每个数据输出一行，即题意中的最小的 t ，如果没有满足要求的 Heron 三角形，输出 -1。

Sample Input

4
1
2
3
4

Sample Output

4
4
4
4

题解

我们可以用海伦公式表示面积

我们设 x = t/2，y = 2S/t，那么

这是pell方程的形式，所以先手算出最小的解 x=2，y=3，然后我们用pell方程的递推式：

我们会发现X增长得很快，到第52个就超过十的三十次方了，因此我们可以先打个表

X[1] = 4;
X[2] = 14;
X[3] = 52;
X[4] = 194;
X[5] = 724;
X[6] = 2702;
X[7] = 10084;
X[8] = 37634;
X[9] = 140452;
X[10] = 524174;
X[11] = 1956244;
X[12] = 7300802;
X[13] = 27246964;
X[14] = 101687054;
X[15] = 379501252;
X[16] = 1416317954;
X[17] = 5285770564;
X[18] = 19726764302;
X[19] = 73621286644;
X[20] = 274758382274;
X[21] = 1025412242452;
X[22] = 3826890587534;
X[23] = 14282150107684;
X[24] = 53301709843202;
X[25] = 198924689265124;
X[26] = 742397047217294;
X[27] = 2770663499604052;
X[28] = 10340256951198914;
X[29] = 38590364305191604;
X[30] = 144021200269567502;
X[31] = 537494436773078404;
X[32] = 2005956546822746114;
X[33] = 7486331750517906052;
X[34] = 27939370455248878094;
X[35] = 104271150070477606324;
X[36] = 389145229826661547202;
X[37] = 1452309769236168582484;
X[38] = 5420093847118012782734;
X[39] = 20228065619235882548452;
X[40] = 75492168629825517411074;
X[41] = 281740608900066187095844;
X[42] = 1051470266970439230972302;
X[43] = 3924140458981690736793364;
X[44] = 14645091568956323716201154;
X[45] = 54656225816843604128011252;
X[46] = 203979811698418092795843854;
X[47] = 761263020976828767055364164;
X[48] = 2841072272208896975425612802;
X[49] = 10603026067858759134647087044;
X[50] = 39571031999226139563162735374;
X[51] = 147681101929045799118003854452;
X[52] = 551153375716957056908852682434;
X[53] = 2056932400938782428517406875284; // 此处就超过 10^30 了

然后就二分判断就过了。

这题根本就不存在 -1。

ＣＯＤＥ

zxy tql %%%%%%

#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#include<bitset>
#include<ctime>
#include<iostream>
#define MAXN 2005
#define LL long long
#define ULL unsigned LL
#define rg register
#define lowbit(x) (-(x) & (x))
#define ENDL putchar('\n')
#define DB double
//#define bs bitset<1005>
//#pragma GCC optimize(2)
//#pragma G++ optimize(3)
//#define int LL
using namespace std;
char char_read_before = 1;
inline int read() {
	int f = 1,x = 0;char s = char_read_before;
	while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x = x * 10 - '0' + s;s = getchar();}
	char_read_before = s; return x * f;
}
inline char readchar() {
	char s = char_read_before;
	while(s == 1 || s == ' ' || s == '\n') s = getchar();
	char_read_before = 1; return s;
}
LL zxy = 100000000;
int n,m,i,j,s,o,k;
DB bg;inline DB Time() {return DB(clock() - bg) / CLOCKS_PER_SEC;}
struct Num{
	LL s[4];
	Num(){s[0]=s[1]=s[2]=s[3]=0;}
	Num(int b) {s[0] = b;s[1] = s[2] = s[3] = 0;}
	Num operator = (int b) {
		s[0] = b;
		s[1] = s[2] = s[3] = 0;
		return *this;
	}
	void tl() {
		s[0] *= 10;
		s[1] = s[1] * 10 + s[0] / zxy;
		s[2] = s[2] * 10 + s[1] / zxy;
		s[3] = s[3] * 10 + s[2] / zxy;
		s[0] %= zxy;
		s[1] %= zxy;
		s[2] %= zxy;
	}
};
inline Num operator *(Num a,Num b) {
	Num c;
	for(int i = 0;i < 4;i ++) {
		LL m = 0;
		for(int j = 0;i+j < 4;j ++) {
			c.s[i+j] += a.s[i] *1ll* b.s[j] + m;
			m = c.s[i+j] / zxy;
			c.s[i+j] %= zxy;
		}
	}return c;
}
inline Num operator +(Num a,Num b) {
	LL m = 0;
	for(int i=0;i<4;i++) {
		a.s[i] += b.s[i] + m;
		m = a.s[i] / zxy;
		a.s[i] %= zxy;
	}return a;
}
inline bool operator < (Num a,Num b) {
	if(a.s[3] != b.s[3]) return a.s[3] < b.s[3];
	if(a.s[2] != b.s[2]) return a.s[2] < b.s[2];
	if(a.s[1] != b.s[1]) return a.s[1] < b.s[1];
	return a.s[0] < b.s[0];
}
inline bool operator >= (Num a,Num b) {return !(a < b);}
inline void print(Num a) {
	int le = 0;
	if(a.s[3]) le = 3;
	else if(a.s[2]) le = 2;
	else if(a.s[1]) le = 1;
	printf("%d",a.s[le]);
	while(le --) printf("%08d",a.s[le]);
	return ;
}
inline Num readn() {
	int f = 1;Num x(0);char s = char_read_before;
	while(s < '0' || s > '9') {if(s == '-') f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x.tl();x = x + Num(s - '0');s = getchar();}
	char_read_before = s; return x;
}
struct mat{
	int n,m;
	Num s[3][3];
	mat(){n=m=0;s[1][1]=s[1][2]=s[2][1]=s[2][2]=Num();}
}A,B;
inline mat operator * (mat a,mat b) {
	mat c; c.n = a.n;c.m = b.m;
	for(int i=1;i<=c.n;i++)
		for(int k=1;k<=a.m;k++)
			for(int j=1;j<=c.m;j++)
				c.s[i][j] = c.s[i][j] + a.s[i][k] * b.s[k][j];
	return c;
}
Num as[100];
signed main() {
	bg = clock();
	A.n = 1;
	A.m = B.n = B.m = 2;
	A.s[1][1] = 2;
	A.s[1][2] = 3;
	B.s[1][1] = 2;
	B.s[2][1] = 1;
	B.s[1][2] = 3;
	B.s[2][2] = 2;
	for(int i = 0;i <= 52;i ++) {
		as[i] = A.s[1][1] * Num(2);
		A = A * B;
	}
	int T = read();
	while(T --) {
		Num nn = readn();
		int l = 0,r = 52,mid;
		while(l < r) {
			mid = l + r >> 1;
			if(as[mid] >= nn) r = mid;
			else l = mid+1;
		}
		print(as[l]);
		ENDL;
	}
	return 0;
}