2.答案是C。代码:
program tou;
var t:char;
begin
for t:='a' to 'd' do
if ord(t<>a)+ord(t=c)+ord(t=d)+ord(t<>d)=1 then writeln(t);
readln;
end.
解释:
ord(true)=1
ord(false)=0,这样一来, ord(t<>'a')+ord(t='c')+ord(t='d')+ord(t<>'d')通过循环得到的thief的结果如果相加等于3
1. 代码:
program tongji;
var i,s,n:integer;
t:char;
begin
readln(n);
for i:=1 to n do
begin
readln(t);
if t='a' then inc(a)
else if t='b' then inc(b)
else if t='c' then inc(c)
else if t='d' then inc(d)
else inc(s);
end;
writeln('A:',a,' ','B:',b,' ','C:',' ','D:',' ');
writeln('qi quan: ',s); {这行输弃权票}
end;