设三个数知分别是x,y,z
10<=x<=99
100<=y<=999
1000<=z<=9999
那么x+y+z=2011的解的个数就是满足要道求的所有算式的个数
令x=10+t1
y=100+t2
z=100+t3
0<=t1<=89，而t1+t2+t3=901
枚榉t1，t2+t3=901-t1，由于0<=t2<=899，
所以当版t1=0~权2:
t2=0~899,有899*2=1798种
当t1=3~89：t2=0～901-t1，有812+813+。。。+898种
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