cicion虛擬貨幣
1. 求C語言寫的高精度除法代碼！！！！！！
這個不是C語言。是pas語言。。。。不過看看把，差不多。
其實這個演算法不難，就是數組模擬除法操作，稍微想像很容易的。Acm比賽中的基本演算法。
我沒有寫這個程序，從信息初學者那裡找了兩個演算法，pascal寫的，根據你用的語言自己轉化下吧，以前用c++寫的找不到了：（
除法 1 : 高精度數 / 單精度數
{
Author : tenshi
Date : 2002-03-10
Problem : High Precision --- Divide
Algorithm : simple
Input : two lines,
1st line : a positive HighPrecision Number
2nd line : a positive Low Precicion Number
Output : times & rest of the two Numbers
}
program HighPrecision3_Multiply1;
const
fn_inp='hp5.inp';
fn_out='hp5.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x,y:hp;
z,w:integer;
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
i:integer;
begin
assign(input,fn_inp);
reset(input);
readln(st);
x.len:=length(st);
for i:=1 to x.len do { change string to HP }
x.s[i]:=ord(st[x.len+1-i])-ord('0');
readln(z);
close(input);
end;
procere Divide(a:hp;b:integer;var c:hp;var d:integer);
{ c:=a div b ; d:=a mod b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
d:=0;
for i:=len downto 1 do { from high to low }
begin
d:=d*10+a.s[i];
c.s[i]:=d div b;
d:=d mod b;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x,z,y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
writeln(w);
close(output);
end;
begin
init;
main;
out;
end.
除法 2 : 高精度數 / 高精度數
{
Author : tenshi
Date : 2002-03-10
Problem : High Precision --- Divide
Algorithm : simple
Input : two lines, each line a positive HighPrecision Number
Output : quotient & molus of the two HighPrecision Numbers
}
program HighPrecision4_Multiply2;
const
fn_inp='hp6.inp';
fn_out='hp6.out';
maxlen=100; { max length of the number }
type
hp=record
len:integer; { length of the number }
s:array[1..maxlen] of integer
{ s[1] is the lowest position
s[len] is the highest position }
end;
var
x:array[1..2] of hp;
y,w:hp; { x:input ; y:output }
procere PrintHP(const p:hp);
var i:integer;
begin
for i:=p.len downto 1 do write(p.s[i]);
end;
procere init;
var
st:string;
j,i:integer;
begin
assign(input,fn_inp);
reset(input);
for j:=1 to 2 do
begin
readln(st);
x[j].len:=length(st);
for i:=1 to x[j].len do { change string to HP }
x[j].s[i]:=ord(st[x[j].len+1-i])-ord('0');
end;
close(input);
end;
procere Subtract(a,b:hp;var c:hp); { c:=a-b, suppose a>=b }
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
for i:=1 to len do { subtract from low to high }
begin
inc(c.s[i],a.s[i]-b.s[i]);
if c.s[i]<0 then
begin
inc(c.s[i],10);
dec(c.s[i+1]); { add 1 to a higher position }
end;
end;
while(len>1) and (c.s[len]=0) do dec(len);
c.len:=len;
end;
function Compare(const a,b:hp):integer;
{
1 if a>b
0 if a=b
-1 if a<b
}
var len:integer;
begin
if a.len>b.len then len:=a.len { get the bigger length of a,b }
else len:=b.len;
while(len>0) and (a.s[len]=b.s[len]) do dec(len);
{ find a position which have a different digit }
if len=0 then compare:=0 { no difference }
else compare:=a.s[len]-b.s[len];
end;
procere Multiply10(var a:hp); { a:=a*10 }
var i:Integer;
begin
for i:=a.len downto 1 do
a.s[i+1]:=a.s[i];
a.s[1]:=0;
inc(a.len);
while(a.len>1) and (a.s[a.len]=0) do dec(a.len);
end;
procere Divide(a,b:hp;var c,d:hp); { c:=a div b ; d:=a mod b }
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a.len;
fillchar(d,sizeof(d),0);
d.len:=1;
for i:=len downto 1 do
begin
Multiply10(d);
d.s[1]:=a.s[i]; { d:=d*10+a.s[i] }
{ c.s[i]:=d div b ; d:=d mod b; }
{ while(d>=b) do begin d:=d-b;inc(c.s[i]) end }
while(compare(d,b)>=0) do
begin
Subtract(d,b,d);
inc(c.s[i]);
end;
end;
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end;
procere main;
begin
Divide(x[1],x[2],y,w);
end;
procere out;
begin
assign(output,fn_out);
rewrite(output);
PrintHP(y);
writeln;
PrintHP(w);
writeln;
close(output);
end;
begin
init;
main;
out;
end.
2. .星期五下午Cici可以回家,通常要花大約三個半小時才能到家的英語
cici can go home on Friday,it usually takes her 3.5 hours to get to her home.
3. ACCESS 用sql語句：例如：alter table時 如何 為欄位 指定 精度 數值范圍
access 應該是不支持 numeric(m,n)的數據類型。
4. 關於西班牙語的間接賓語
1. nos就是賓格代詞所謂直接賓語出現，convenia的主語是ejecución
2. 是自復動詞的用法（台灣的書好像稱之反身動詞）即quedarse相當於estar，多問問你的老師，畢竟口頭的傳授勝過這種筆頭的解釋。
另外你這兩句似乎都有點小錯誤啊，既然求教是否應當持有一個嚴謹的態度，這樣也許早就有人在我之前答復您了~
5. 銀河護衛隊星爵要阻止羅南摧毀山達爾星之前整理裝備的背景音樂，唱的好像是「cicicicicicic
1. Blue Swede - Hooked on a Feeling
2. Raspberries - Go All the Way
3. Norman Greenbaum - Spirit in the Sky
4. David Bowie - Moonage Daydream
5. Elvin Bishop - Fooled Around and Fell in Love
6. 10Cc - I'm Not in Love
7. Jackson 5 - I Want You Back
8. Redbone - Come and Get Your Love
9. The Runaways - Cherry Bomb
10. Rupert Holmes - Escape (the Pina Colada Song)
11. The Five Stairsteps - O-O-H Child
12. Marvin Gaye/Tammi Terrell - Ain't No Mountain High Enough