SGU 106 模线性方程

最新推荐文章于 2016-05-27 02:10:28 发布

原创最新推荐文章于 2016-05-27 02:10:28 发布 · 1.2k 阅读

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本文介绍了一种解决特定线性方程组问题的算法实现细节，包括特殊情况处理及一般情况下的求解过程。通过扩展欧几里得算法求解未知数，并结合边界条件确定解的存在性和范围。

一气之下变量全变成__int64就A了，不知道为什么要会越界。
先考虑一些特殊情况。
其它情况解两个方程组即可
x1 <= x0 + k*a <= x2;

y1 <= y0 - k*b <= y2;

求出来的k值取相交部分。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL __int64
// x1 = y2,  y1 = x2 - (a/b) * y2;
LL ex_gcd(LL a, LL b, LL &x, LL &y) {
	if (!b) {
		x = 1;
		y = 0;
		return a;
	}
	LL t = ex_gcd(b, a % b, x, y);
	LL xx = x;
	x = y;
	y = xx - a / b * y;
	return t;
}

int main() {
	LL i;
	int cas;
	LL a, b, c, x, y, x1, x2, y1, y2;
	cin >> a >> b >> c >> x1 >> x2 >> y1 >> y2;
	if (a < 0)
		a = -a, b = -b;
	else
		c = -c;
	if (!a && !b && c) {
		puts("0");
		return 0;
	}
	if (!a && !b && !c) {
		printf("%I64d", (x2 - x1 + 1) * (y2 - y1 + 1));
		return 0;
	}
	if (!a && b) {
		LL ty = c / b;
		if (c % b == 0 && ty >= y1 && ty <= y2)
			printf("%I64d\n", (x2 - x1 + 1));
		else
			puts("0");
		return 0;
	}
	if (a && !b) {
		LL tx = c / a;
		if (c % a == 0 && tx >= x1 && tx <= x2)
			printf("%I64d\n", (y2 - y1 + 1));
		else
			puts("0");
		return 0;
	}
	LL t = ex_gcd(a, b, x, y);
	if (c % t != 0) {
		puts("0");
		return 0;
	}
	a /= t;
	b /= t;
	c /= t;
	x *= c;
	y *= c;
	//	printf("a = %d b = %d c = %d x = %d y = %d\n", a, b, c, x, y);
	LL l1, r1;
	x1 -= x;
	x2 -= x;
	if (b < 0) {
		b = -b;
		x1 = -x1;
		x2 = -x2;
		swap(x1, x2);
	}
	if (x1 >= 0)
		l1 = (x1 + b - 1) / b;
	else
		l1 = x1 / b;

	if (x2 >= 0)
		r1 = x2 / b;
	else {
		LL t = x2 / b;
		LL tp = x2 - t * b;
		if (tp < 0)
			t--;
		r1 = t;
	}

	LL l2, r2;
	y1 -= y;
	y2 -= y;
	y1 = -y1;
	y2 = -y2;
	swap(y1, y2);
	if (a < 0) {
		a = -a;
		y1 = -y1;
		y2 = -y2;
		swap(y1, y2);
	}

	if (y1 >= 0)
		l2 = (y1 + a - 1) / a;
	else
		l2 = y1 / a;

	if (y2 >= 0)
		r2 = y2 / a;
	else {
		LL t = y2 / a;
		LL tp = y2 - t * a;
		if (tp < 0)
			t--;
		r2 = t;
	}
	//	printf("l1 = %d r1 = %d l2 = %d r2 = %d\n", l1, r1, l2, r2);
	l1 = max(l1, l2);
	r1 = min(r1, r2);
	if (l1 > r1)
		puts("0");
	else
		printf("%I64d\n", r1 - l1 + 1);
	return 0;
}