spoj 6 Simple Arithmetics
这道题很烦。。我的代码也很烦。。我想也没人愿意会看
下面是为我找错误提供方便的几个地方：
http://blog.csdn.net/tiaotiaoyly/archive/2008/02/11/2087975.aspx
他总结得很好。。但是对我帮助不大。。呵呵。。不过可以参考一下
开始几个错误还是自己的数据测出来的。。
我最致命的错误是横线的输出
横线的长度其实取决于上下两个数的长度的。。我一直以为是取决于最长数的长度
以下是代码。。刚好利用这个题熟悉了一下大数模板
#include
<
iostream
>
#define MaxN
3500
#define max(a,b) (a
>
b?a:b)#define max3(a,b,c) (max(a,max(b,c)))char str[MaxN],x[MaxN],y[MaxN];struct HP{
int
len
,s[MaxN];}a,b,c;void StrHP(
const
char
*
s,HP
&
x){ x.len
=
(
int
)strlen(s);
for
(
int
i
=
1
;i
<=
x.len;i
++
) x.s[i]
=
s[x.len
-
i]
-
'
0';
}void Int2HP(
int
inte,HP
&
x){
if
(!inte) { x.len
=
1
; x.s[
1
]
=
0
; return; } memset(x.s,
0
,sizeof(x.s));
for
(x.len
=
0
;inte
>
0
;) { x.s[
++
x.len]
=
inte%
10
; inte
/=
10
; }}void Multi(
const
HP a,
const
HP b,HP
&
c){
int
i,j; c.len
=
a.len
+
b.len;
for
(i
=
1
;i
<=
c.len;i
++
) c.s[i]
=
0
;
for
(i
=
1
;i
<=
a.len;i
++
)
for
(j
=
1
;j
<=
b.len;j
++
) c.s[i
+
j
-
1
]
+=
a.s[i]
*
b.s[j];
for
(i
=
1
;i
<
c.len;i
++
) { c.s[i
+
1
]
+=
c.s[i]
/
10
; c.s[i]%
=
10
; }
while
(c.s[i]) { c.s[i
+
1
]
=
c.s[i]
/
10
; c.s[i]%
=
10
; i
++
; }
while
(i
>
1
&&
!c.s[i])i
--
; c.len
=
i;}void Plus(
const
HP a,
const
HP b,HP
&
c){
int
i; c.s[
1
]
=
0
;
for
(i
=
1
;i
<=
a.len||i
<=
b.len||c.s[i];i
++
) {
if
(i
<=
a.len) c.s[i]
+=
a.s[i];
if
(i
<=
b.len) c.s[i]
+=
b.s[i]; c.s[i
+
1
]
=
c.s[i]
/
10
; c.s[i]%
=
10
; } c.len
=
i
-
1
;
if
(c.len
==
0
) c.len
=
1
;}void Substract(
const
HP a,
const
HP b,HP
&
c){
for
(
int
i
=
1
,j
=
0
;i
<=
a.len;i
++
) { c.s[i]
=
a.s[i]
-
j;
if
(i
<=
b.len) c.s[i]
-=
b.s[i];
if
(c.s[i]
<
0
) { j
=
1
; c.s[i]
+=
10
; }
else
j
=
0
; } c.len
=
a.len;
while
(c.len
>=
1
&&
!c.s[c.len])c.len
--
;
if
(!c.len)c.len
=
1
;}void PrintHP(HP
&
x){
for
(
int
i
=
x.len;i
>=
1
;i
--
) printf(
"
%d
"
,x.s[i]); }
int
main(){
//
freopen(
"
1.out
"
,
"
w
"
,stdout);
int
T,
len
,len2,i,j; char
*
ptr; char oper; HP tmp,tmp2; scanf(
"
%d
"
,
&
T);
while
(T
--
) { len2
=
len
=-
1
; scanf(
"
%s
"
,str);
for
(ptr
=
x,i
=
j
=
0
;str[i];i
++
) {
if
(str[i]
>=
'
0'&&str[i]<='9')
ptr[j
++
]
=
str[i];
else
{ oper
=
str[i]; ptr[j]
=
'
\0';
j
=
0
; ptr
=
y; } } ptr[j]
=
'
\0';
StrHP(x,a); StrHP(y,b); switch(oper) {
case
'
+':
case
'
-':
if
(oper
==
'
+')Plus(a,b,c);
else
Substract(a,b,c);
len
=
max3(a.len,b.len
+
1
,c.len);
//
带个符号 len2
=
max(b.len
+
1
,c.len);
for
(i
=
0
;i
<
len
-
a.len;i
++
) putchar(
'
');
PrintHP(a);
//
不带回车的print putchar(
'
\n');
for
(i
=
0
;i
<
len
-
b.len
-
1
;i
++
) putchar(
'
');
putchar(oper); PrintHP(b); putchar(
'
\n');
for
(i
=
0
;i
<
len
;i
++
) i
<
len
-
len2?putchar(
'
'):putchar('-');
putchar(
'
\n');
for
(i
=
0
;i
<
len
-
c.len;i
++
) putchar(
'
');
PrintHP(c); putchar(
'
\n');
break;
case
'
*':
Multi(a,b,c);
len
=
max3(a.len,b.len
+
1
,c.len);
for
(i
=
0
;i
<
len
-
a.len;i
++
) putchar(
'
');
PrintHP(a); putchar(
'
\n');
for
(i
=
0
;i
<
len
-
b.len
-
1
;i
++
) putchar(
'
');
putchar(
'
*');
PrintHP(b); putchar(
'
\n');
Int2HP(b.s[
1
],tmp); Multi(a,tmp,tmp2); len2
=
max(b.len
+
1
,tmp2.len);
for
(i
=
0
;i
<
len
;i
++
) i
<
len
-
len2?putchar(
'
'):putchar('-');
putchar(
'
\n');
for
(i
=
0
;i
<
len
-
tmp2.len;i
++
) putchar(
'
');
PrintHP(tmp2); putchar(
'
\n');
for
(i
=
2
;i
<=
b.len;i
++
) { Int2HP(b.s[i],tmp); Multi(a,tmp,tmp2);
for
(j
=
0
;j
<
len
-
tmp2.len
-
i
+
1
;j
++
) putchar(
'
');
PrintHP(tmp2); putchar(
'
\n');
}
if
(b.len
>
1
) { len2
=
max(tmp.len
+
b.len
-
1
,c.len);
for
(i
=
0
;i
<
len
;i
++
) i
<
len
-
len2?putchar(
'
'):putchar('-');
putchar(
'
\n');
for
(i
=
0
;i
<
len
-
c.len;i
++
) putchar(
'
');
PrintHP(c); putchar(
'
\n');
} break; } putchar(
'
\n');
} return
0
;}