Hsuan 发表于 2014-9-8 22:18 能讲下思路和注意?好容易做错 |
本帖最后由 Hsuan 于 2014-9-8 22:22 编辑 没想到什么特别好的方法。$z+y<= 16 ;0<=y<=3 ;0<= z-3y<=8 ;$所以,$\frac{z-8}{3}<=y ,z+\frac{z-8}{3}<= z+y<=16 ,z<=14.$此时,$y=\frac{z-8}{3}=2.$ |
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