算法笔记之旅——百鸡问题时间超限

一开始是用最简单的三层循环，每个x,y,z一个个遍历，符合条件即输出，时间超限：

		for (int x = 0; x <= 100; x++) {
			for (int y = 0; y <= 100; y++) {
				for (int z = 0; z <= 100; z++) {
					if (x + y + z == 100 && 5 * x + 3 * y + (1.0 / 3)*z <= n) {
						cout << "x=" << x << ",y=" << y << ",z=" << z << endl;
					}
				}
			}
		}

考虑通过x得出y的范围，y<=100-x，z=100-x-y.

for (x = 0; x <= 100; x++)
			for (y = 0; y <= 100 - x; y++)
			{
				z = 100 - x - y;
				if (x*5.0 + y * 3.0 + z / 3.0 <= n * 1.0)
					cout << "x=" << x << ",y=" << y << ",z=" << z << endl;
			}