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atttx123 2009-01-17 06:51:00
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Program asd;

Var n,i,j,p,q: longint;

Begin

p := 1;

q := 0;

read(n);

If n=1 Then

Begin

writeln(1);

halt;

End;

For i:=1 To n Do

Begin

j := i;

while j mod 2 = 0 Do

Begin

inc(q);

j := j Div 2;

End;

while j mod 5 = 0 Do

Begin

dec(q);

j := j Div 5;

End;

q := q Mod 4;

while q<0 Do inc(q,4);

p := p*j Mod 10;

End;

p := p*6 Mod 10;

For i:=1 To q Do

p := p*2 Mod 10;

Case p of

1 : writeln(’drxh’);

2 : writeln(’dyf’);

3 : writeln(’hades’);

4 : writeln(’xiaokeke’);

5 : writeln(’xiaokaka’);

6 : writeln(’atttx123’);

7 : writeln(’h’);

8 : writeln(’LIFE’);

9 : writeln(’dadi’);

End;

End.

#1 Zx.MYS@2008-08-21 02:48:00
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10^100

如何read(n)?

#2 LIFE@2008-10-29 03:44:00
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用字符串

#3 cc@2008-10-29 03:55:00
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ye
#4 why19931123@2009-01-16 05:05:00
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数据太弱了!!!

最末非0原程序A了(只改了输出)!!!

var

s,o,p,m,n:longint;

begin

readln(m);

o:=1;

for n:=1to m do

begin

p:=n;

while(p mod 2=0)do

begin

inc(s);

p:=p div 2;

end;

while(p mod 5=0)do

begin

dec(s);

p:=p div 5;

end;

o:=(o*p)mod 10;

end;

o:=(o shl(s and 3+4)mod 10);

case o of

1:write(’drxh’);

2:write(’dyf’);

3:write(’hades’);

4:write(’xiaokeke’);

5:write(’xiaokaka’);

6:write(’atttx123’);

7:write(’h’);

8:write(’LIFE’);

9:write(’dadi’);

end;

end.

#5 Mato完整版@2009-01-17 06:51:00
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最后的输出还可以减一点,

其实"hades","xiaokaka","h"和"dadi"是不可能获胜的。

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