y+f=20 (1)s+f=11 (2)y+f+s=27 (3)由(1)+(2)可得：(y+f)+(s+f)=20+11即y+f+s+f=31 (4)由(4)-(3)可得：(y+f+s+f)-(y+f+s)=31-27即f=4 (5)由(1)-(5)可得：y=16由(2)-(5)可得：s=7圆圈、三角、方块分别为16，7，4
