UVA 618 - Doing Windows（数论）

题目链接：618 - Doing Windows

题意：给定一个大小不能变的屏幕，和四个大小可以变的窗口，变化要保持长宽比，问这四个窗口能不能调整后全部放下正好填满屏幕，不能重叠

思路：情况一共就几种：4个叠一起，3个叠一起+一个，2个和2个，一个和两个叠一起在一个，把这几种情况全判断了就可以了，判断过程利用gcd，lcm可以求边长。

代码：

#include <stdio.h>
#include <string.h>

long long gcd(long long a, long long b) {
    if (b == 0)
	return a;
    return gcd(b, a % b);
}

long long lcm(long long a, long long b) {
    return a / gcd(a, b) * b;
}

struct Win {
    long long x, y;
    bool scanf_() {
	scanf("%lld%lld", &x, &y);
	if (x == 0 || y == 0)
	    return false;
	long long t = gcd(x, y); x /= t; y /= t;
	return true;
    }
    void swap() {
	long long t = x;
	x = y;
	y = t;
    }
} win, w[4];

bool check1(Win a, Win b, Win c, Win d) {
    long long sum = 0;
    if (win.y % a.y)
	return false;
    sum += win.y / a.y * a.x;
    if (win.y % b.y)
	return false;
    sum += win.y / b.y * b.x;
    if (win.y % c.y)
	return false;
    sum += win.y / c.y * c.x;
    if (win.y % d.y)
	return false;
    sum += win.y / d.y * d.x;
    if (sum != win.x)
	return false;
    return true;
}

bool check(Win a, Win b, Win c, long long wx, long long wy) {
    if (wy <= wx / a.x * a.y) return false;
    long long yy = wy - wx / a.x * a.y;
    long long sum = 0;
    if (yy % b.y) return false;
    sum += yy / b.y * b.x;
    if (yy % c.y) return false;
    sum += yy / c.y * c.x;
    if (sum != wx) return false;
    return true;
}

bool judge3(Win a, Win b, Win c, long long wx, long long wy) {
    if (check(a, b, c, wx, wy)
	    || check(b, a, c, wx, wy)
	    || check(c, a, b, wx, wy)) return true;
    return false;
}

bool judge2(Win a, Win b, Win c, long long wx, long long wy) {
    if (judge3(a, b, c, wx, wy))
	return true;
    a.swap();
    b.swap();
    c.swap();
    if (judge3(a, b, c, wy, wx))
	return true;
    return false;
}

bool check2(Win a, Win b, Win c, Win d) {
    long long sum = 0;
    if (win.x % d.x)
	return false;
    long long wy = win.x / d.x * d.y;
    if (wy >= win.y)
	return false;
    long long yy = win.y - wy;
    if (judge2(a, b, c, win.x, yy))
	return true;
    if (yy % a.y) return false;
    sum += yy / a.y * a.x;
    if (yy % b.y) return false;
    sum += yy / b.y * b.x;
    if (yy % c.y) return false;
    sum += yy / c.y * c.x;
    int flag = 0;
    if (sum == win.x) flag = 1;
    return flag;
}

bool check3(Win a, Win b, Win c, Win d) {
    long long sum = 0;
    long long wy = lcm(a.y, b.y);
    if (wy >= win.y)
	return false;
    sum += wy / a.y * a.x;
    sum += wy / b.y * b.x;
    if (win.x % sum)
	return false;
    if (win.y <= win.x / sum * wy)
	return false;
    long long yy = win.y - win.x / sum * wy;
    sum = 0;
    if (yy % c.y)
       	return false;
    sum += yy / c.y * c.x;
    if (yy % d.y)
	return false;
    sum += yy / d.y * d.x;
    if (win.x != sum)
	return false;
    return true;
}

bool judge() {
    if (check1(w[0], w[1], w[2], w[3]))
	return true;
    if (check2(w[0], w[1], w[2], w[3])
	    || check2(w[0], w[1], w[3], w[2])
	    || check2(w[0], w[3], w[2], w[1])
	    || check2(w[3], w[1], w[2], w[0]))
       	return true;
    if (check3(w[0], w[1], w[2], w[3])
	    || check3(w[0], w[2], w[1], w[3])
	    || check3(w[0], w[3], w[1], w[2]))
	return true;
    return false;
}

bool solve() {
    if (judge())
	return true;
    win.swap();
    for (long long i = 0; i < 4; i++)
	w[i].swap();
    if (judge())
	return true;
    return false;
}

int main() {
    long long cas = 0;
    while (~scanf("%lld%lld", &win.x, &win.y) && win.x || win.y) {
	for (long long i = 0; i < 4; i++)
	    w[i].scanf_();
	printf("Set %lld: %s\n", ++cas, solve()? "Yes":"No");
    }
    return 0;
}

UVA 618 - Doing Windows（数论）,码迷,mamicode.com