[U]3.2.4 Feed Ratios 枚举
简答的枚举题，其他的不多说了.... 切掉之后看了下题解，复习了一遍高斯消元法和克莱姆法则。发现还是数学方法好
简答的枚举题，其他的不多说了....
切掉之后看了下题解，复习了一遍高斯消元法和克莱姆法则。发现还是数学方法好啊。
虽然枚举很省coding时间,于是乎，抱着节省code时间的态度,决定开始用模板类.....
克莱姆法则有几条重要的
1.非齐次线性方程，系数矩阵D=0时，有无穷多解。D≠0时，有唯一解。
2.齐次线性方程.系数矩阵D=0时，有解。否则无解。
另外克莱姆法则的时间效率低，因为行列式的运算=.=
使用高斯消元法，将行列式变成上下三角行列式，接着化系数，变成对角矩阵。就可以看出解了。方便解小数解。
Code可以无视:
/*
ID:sevenst4
LANG:C++
PROG:ratios
*/
#include<stdio.h>
using namespace std;
int main()
{
freopen( "ratios.in","r",stdin );
freopen( "ratios.out","w",stdout );
int x[4],y[4],z[4];
for( int i=0;i<br>
<br>
<p><br>
</p>
</stdio.h>
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