assembly language implementation from 1 to 100 (1+2+...+100)

Implement 1+2+...+100 with assembly language

;Classroom Assignments;Calculate 1+2+...+100DATA SEGMENT COUNT DW0    ;CountDATA ENDS stack SEGMENT PARA stack BUF DW 20H DUP (0) LEN EQU $-bufstack endscodeseg segmentassumeCS:CODESEG,SS:STACK,DS:DATASTART:;Initialize stack segment SS and data segment DS  MOVAx,stackMOVSs,axMOVSp,lenMOVAx,dataMOVDs,axMOVcx,64h;Cycle 100 times  MOVAx0S:    ADDcount,01hADDax,count LOOP SMOVDl,alMOVAl,ahPagerDisp_2_hexMOVAL,DLPagerDisp_2_hexPagerDisp_crefMOVah,4chINT21H;outputs the high four bits of Al and the lower four bits respectivelyDisp_2_hex:  PUSHAXPUSHBXPUSHF  MOVAH,0 ;Qing 0  MOVbl,10h;for Division  DIVBL;AL: Quotient high AH remainder low  PagerDisp_1_hex;results of output Al  MOVAl,ahPagerDisp_1_hexPopf  POPBXPOPAXRET;output The numbers and letters of AlDisp_1_hex:    PUSHAXPUSHDXPUSHF  MOVDl,alCMPDl the  JbeL_1;less than or equal to 9 skips the next statement  ADDdl,27h;greater than 10 is converted to lowercase letters and executed to the end of the RET to the following L_1                ;The Number 10 and the character a difference is (27H);Digital output (ASCII code of 1 is 31H)L_1:    ADDdl,30h;convert numbers to characters  MOVah,02hINT21HPopf  POPDXPOPAXRET;output carriage return line breakDisp_cref:  PUSHDXPUSHAXMOVah,02hMOVDL,0DHINT21HMOVdl,0ahINT21HPOPAXPOPDXRETcodeseg ENDSEND START

Output is 13ba (hex of 5050)

assembly language implementation from 1 to 100 (1+2+...+100)