1樓:哀傷_無限
答案不唯一,abc/ab/ac都是答案
vari1,i2,i3,j:longint;
begin
for i1:=1 to 2 do
for i2:=1 to 2 do
for i3:=1 to 2 do
if ord(i1=1)+ord(i1=1)+ord((i3=2) and (i2=2))=2 then begin
if i1=1 then write('a ');
if i2=1 then write('b ');
if i3=1 then write('c ');
writeln;
end;
end.
原理:列舉每個人吃或沒吃的情況,1是吃了,b是沒吃,之後,因為ord(true)=1 ord(false)=0,判斷出說真話的有兩人,之後輸出即可。
樓主!手打的!!!!!
2樓:匿名使用者
vari1,i2,i3,j:longint;
begin
for i1:=1 to 2 do
for i2:=1 to 2 do
for i3:=1 to 2 do
if ord(i1=1)+ord(i1=1)+ord((i3=2) and (i2=2))=2 then begin
if i1=1 then write('a ');
if i2=1 then write('b ');
if i3=1 then write('c ');
writeln;
end;
end.
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