GMOJ7177 鱼跃龙门题解

GMOJ 7177 鱼跃龙门 题解

题面

显然是求最小的x 使得$$\frac{x\times (x+1)}{2}\equiv 0(modn)$$
即$$x\times (x+1)\equiv 0(mod2n)$$
令$$\begin{gathered} d=gcd(x,2n), \\ x=x_{1}d, \\ 2n=n_{1}d. \end{gathered}$$
则$$x_{1}d\times (x_{1}d+1)\equiv 0(mod{n}_{1}d)$$
即$$x_1\times (x_{1}d+1)\equiv 0(mod{n}_1)$$
注意到$$gcd(x_1,n_1)=1$$
故$$x_{1}d+1\equiv 0(mod{n}_1)$$
转换形式. 可得$$-x_{1}d+k{n}_1=1$$
视$d$为常量,可用扩展欧几里得算法求得一个小于$0$的$-x_1$
因此最小的$x=x_{1}d$
枚举$d$即可.

#include <cstdio>
#include <iostream>
#include <cstring>
#define R register
using namespace std;
typedef long long LL;
const LL inf = 1e15;
LL x, yy, T, n, y[200000][2], cntP, cntY, vis[1000100], p[1000000];
LL ans = inf;
inline LL read() {
	LL res; char ch = getchar();
	for(; ch > '9' || ch < '0'; ch = getchar());
	for(res = 0; ch >= '0' && ch <= '9'; res = res * 10 + ch - '0', ch = getchar());
	return res;
}
inline LL gcd(LL a, LL b) {return b == 0 ? a : gcd(b, a % b);}
void exgcd(LL a, LL b) {
	if(b == 0) {
		x = 1, yy = 0;
		return ;
	}
	exgcd(b, a % b);
	LL tmp = x;
	x = yy;
	yy = tmp - a / b * yy;
	return ;
}
void check(LL now) {
	if(gcd(now, n / now) > 1) return ;
	exgcd(now, n / now);
	if(x >= 0) x = x % (n / now) - (n / now);
	x = -x * now;
	ans = min(ans, x);
}
void dfs(int k, LL now) {
	if(k > cntY) {
		check(now);
		return ;
	}
	LL tmp = now;
	for(R int i = 0; i <= y[k][1]; ++i) {
		dfs(k + 1, tmp);
		tmp /= y[k][0];
	}
	return ;
}
int main() {
	for(register int i = 2; i <= 1000000; ++i) {
		if(!vis[i]) p[++cntP] = i;
		for(register int j = 1; j <= cntP; ++j) {
			int tmp = p[j] * i;
			if(tmp > 1000000) break;
			vis[tmp] = 1;
			if(i % p[j] == 0) break;
		}
	}
	T = read();
	for(; T; T--) {
		n = read() * 2;
		LL tmp = n;
		ans = inf;
		memset(y, 0, sizeof(y));
		cntY = 0;
		for(R int i = 1; i <= cntP && p[i] <= tmp; ++i) {
			if(tmp % p[i] == 0) {
				++cntY;
				y[cntY][0] = p[i];
				while(tmp % p[i] == 0) {
					y[cntY][1] ++;
					tmp /= p[i];
				}
			}
		}
		if(tmp > 1) y[++cntY][0] = tmp, y[cntY][1] = 1;
		dfs(1, n);
		printf("%lld\n",ans);
	}
	return 0;
}