Count Me II

Define $N(a,b,c,...)$ is the amount of possibilities of the number pair $(a,b,c,...)$
Let the number be $\overline{ABCDE}$ (any digit can be 0)
When $A=0$ ,consider $(B+C)+(D+E)=18$
If $B+C=0,D+E=18$ ,then $N(B,C)=1,N(D,E)=1$ ,so $N(B,C,D,E)=1$
If $B+C=1,D+E=17$ ,then $N(B,C)=2,N(D,E)=2$ ,so $N(B,C,D,E)=4$
If $B+C=2,D+E=16$ ,then $N(B,C)=3,N(D,E)=3$ ,so $N(B,C,D,E)=9$
...
If $B+C=9,D+E=9$ ,then $N(B,C)=10,N(D,E)=10$ ,so $N(B,C,D,E)=100$
If $B+C=10,D+E=8$ ,then $N(B,C)=9,N(D,E)=9$ ,so $N(B,C,D,E)=81$
...
If $B+C=18,D+E=0$ ,then $N(B,C)=1,N(D,E)=1$ ,so $N(B,C,D,E)=1$
So,when $A=0,N(B,C,D,E)=1+4+9+...+100+81+64+...+4+1=670$

When $A=1$ ,consider $(B+C)+(D+E)=17$
If $B+C=0,D+E=17$ ,then $N(B,C)=1,N(D,E)=2$ ,so $N(B,C,D,E)=2$
If $B+C=1,D+E=16$ ,then $N(B,C)=2,N(D,E)=3$ ,so $N(B,C,D,E)=6$
If $B+C=2,D+E=15$ ,then $N(B,C)=3,N(D,E)=4$ ,so $N(B,C,D,E)=12$
...
If $B+C=8,D+E=9$ ,then $N(B,C)=9,N(D,E)=10$ ,so $N(B,C,D,E)=90$
If $B+C=9,D+E=8$ ,then $N(B,C)=10,N(D,E)=9$ ,so $N(B,C,D,E)=90$
...
If $B+C=17,D+E=0$ ,then $N(B,C)=2,N(D,E)=1$ ,so $N(B,C,D,E)=2$
So,when $A=1,N(B,C,D,E)=2(2+6+12+...+90)=660$

When $A=2,\overline{ABCDE}=20079,20088,20097,20169,20178$ (5 possibilities)
So,there are $670+660+5=1335$ possibilities.