题目
求1000到9999之间所有回文数的个数
提问时间:2020-11-30
答案
set talk off
clear
n=0
for x=1000 to 9999
a=int(x/1000)
b=int((x-a*1000)/100)
c=int((x-a*1000-b*100)/10)
d=mod(x,10)
if a=d and b=c and mod(x,6)=0
n=n+1
endif
endfor
set talk on
return
clear
n=0
for x=1000 to 9999
a=int(x/1000)
b=int((x-a*1000)/100)
c=int((x-a*1000-b*100)/10)
d=mod(x,10)
if a=d and b=c and mod(x,6)=0
n=n+1
endif
endfor
set talk on
return
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