URAL1153 Supercomputer 大数开方模板
今天下午哈尔滨赛区网络预选赛1003出了一道大数开方的题，自己用了XX大学的模板，并错误的深信不疑，导致WA11次，悲剧RANK94
如果1Y了，肯定是另一种结果啊。情何以堪……
赛后得知模板用错，用中山大学的模板区做URAL1153Supercomputer（大数开方），小菜一碟。
题目大意：N=x*(x+1)/2，给你N (N < 10600)，输出x
分析：求sqrt(2*N)即可
代码
#include
<
iostream
>
#include
<
string
>
#include
<
cstdlib
>
#include
<
algorithm
>
using
namespace
std;
#define
MAXN 2000
int
big(
char
s1[],
char
s2[]){
int
len1,len2,i,q;
q
=
0
;
while
(s1[q]
==
'
0
'
) q
++
;
strcpy(s1,s1
+
q);
if
(strlen(s1)
==
0
){
s1[
0
]
=
'
0
'
;
s1[
1
]
=
0
;
}
q
=
0
;
while
(s2[q]
==
'
0
'
) q
++
;
strcpy(s2,s2
+
q);
if
(strlen(s2)
==
0
){
s2[
0
]
=
'
0
'
;
s2[
1
]
=
0
;
}
len1
=
strlen(s1);
len2
=
strlen(s2);
if
(len1
>
len2)
return
1
;
else
if
(len1
<
len2)
return
0
;
else
{
for
(i
=
0
;i
<
len1;i
++
){
if
(s1[i]
>
s2[i])
return
1
;
else
if
(s1[i]
<
s2[i])
return
0
;
}
}
return
0
;
}
void
mul(
char
s[],
int
t,
char
re[]){
//
乘
int
left,i,j,k,len;
char
c;
left
=
0
;
j
=
0
;
for
(i
=
strlen(s)
-
1
;i
>=
0
;i
--
){
k
=
t
*
(s[i]
-
'
0
'
)
+
left;
re[j
++
]
=
(k
%
10
)
+
'
0
'
;
left
=
k
/
10
;
}
while
(left
>
0
){
re[j
++
]
=
(left
%
10
)
+
'
0
'
;
left
/=
10
;
}
re[j]
=
0
;
len
=
strlen(re);
for
(i
=
0
;i
<
len
/
2
;i
++
){
c
=
re[i];
re[i]
=
re[len
-
1
-
i];
re[len
-
1
-
i]
=
c;
}
return
;
}
void
sub(
char
a[],
char
b[]){
//
减
int
left,len1,len2,temp,j;
len1
=
strlen(a)
-
1
;
len2
=
strlen(b)
-
1
;
left
=
0
;
while
(len2
>=
0
){
temp
=
a[len1]
-
b[len2]
+
left;
if
(temp
<
0
){
temp
+=
10
;
left
=-
1
;
}
else
left
=
0
;
a[len1]
=
temp
+
'
0
'
;
len1
--
;
len2
--
;
}
while
(len1
>=
0
){
temp
=
a[len1]
-
'
0
'
+
left;
if
(temp
<
0
){
temp
+=
10
;
left
=-
1
;
}
else
left
=
0
;
a[len1]
=
temp
+
'
0
'
;
len1
--
;
}
j
=
0
;
while
(a[j]
==
'
0
'
) j
++
;
strcpy(a,a
+
j);
if
(strlen(a)
==
0
){
a[
0
]
=
'
0
'
;
a[
1
]
=
0
;
}
return
;
}
void
sqr(
char
s[],
char
re[]){
//
开方
char
temp[MAXN];
char
left[MAXN];
char
p[MAXN];
int
i,j,k,len1,len2,q;
len1
=
strlen(s);
if
(len1
%
2
==
0
){
left[
0
]
=
s[
0
];
left[
1
]
=
s[
1
];
left[
2
]
=
0
;
j
=
2
;
}
else
{
left[
0
]
=
s[
0
];
left[
1
]
=
0
;
j
=
1
;
}
re[
0
]
=
'
0
'
;
re[
1
]
=
0
;
q
=
0
;
while
(j
<=
len1){
mul(re,
20
,temp);
len2
=
strlen(temp);
for
(i
=
9
;i
>=
0
;i
--
){
temp[len2
-
1
]
=
i
+
'
0
'
;
mul(temp,i,p);
if
(
!
big(p,left))
break
;
}
re[q
++
]
=
i
+
'
0
'
;
re[q]
=
0
;
sub(left,p);
len2
=
strlen(left);
left[len2]
=
s[j];
left[len2
+
1
]
=
s[j
+
1
];
left[len2
+
2
]
=
0
;
j
+=
2
;
}
}
int
main(){
char
s[MAXN],s2[MAXN],re[MAXN];
int
an[MAXN];
char
ans[MAXN];
int
i;
while
(scanf(
"
%s
"
,s)
!=
EOF ){
mul(s,
2
,s2);
strcpy(s,s2);
re[
0
]
=
0
;
sqr(s,re);
i
=
0
;
while
(re[i]
==
'
0
'
) i
++
;
strcpy(re,re
+
i);
printf(
"
%s\n
"
,re);
}
return
0
;
}
今天的1003题，用此模板，再加上大整数任意进制转换，可以说是用来秒杀的。
今天RP为何如此低？可悲……
大整数任意进制转换下一篇文章给出