High Speed Trains
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
Submit Statistic Next Problem
Problem Description
The kingdom of Flatland has n cities. Recently the king of Flatland visited Japan and was amazed by high speed trains Shinkansens going all around the country. Therefore he decided to build the system of high speed trains in Flatland.
Each high speed train line will be bidirectional and connect exactly two different cities of Flatland. Although there is actually no need of high speed trains in Flatland, the king ordered that there must be at least one high speed train line from each city of Flatland.
The minister of transportation told the king that there are several train system satisfying his requirements. The king was amazed by the fact and asked the minister to count the number of possible systems.
Help the minister to calculate the number of train systems.
Input
The input file contains one integer number n (2 ≤ n ≤ 100)
Output
Output one integer number — the number of different train systems that can be arranged in Flatland.
Sample Input
4
Sample Output
41
Source
Andrew Stankevich Contest 22
题目大意：
就是你 n 座城市，然后保证每个城市都有一个高铁通往另一个城市（姑且称他为高铁），但是两个城市之间只能有一条铁路,然后让你求的是这样的方法数。
解题思路：
其实我们可以借助容斥原理的思路来做这道题，我们就可以求出总数来，因为一共有 n 座城市，那么最多有sum = n*(n-1)/2条边，然后总数就是2^sum，然后我们可以考虑抛弃一个点，抛弃两个点…一直到抛弃n个点。那么设 ans[n] 为有 n 座城市的方法数，那么我们就可以列出公式：
ans[n]=2sum?ans[n?1]?C(n,1)?ans[n?2]?C(n,2)?...?ans[1]?C(n,n?1)?ans[0]?C(n,n)
注意ans[1]=0,因为我们修一条高铁的时候连接的是两条边，那么不可能出现一个城市的时候。
列出公式来之后就是运算了，这里还有大数运算（还是直接套的模板）
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
#define DIGIT 4 //四位隔开,即万进制
#define DEPTH 10000 //万进制
#define MAXX 1000+5 //题目最大位数/4，要不大直接设为最大位数也行
typedef int bignum_t[MAXX+1];
/************************************************************************/
/* 读取操作数，对操作数进行处理存储在数组里 */
/************************************************************************/
int read(bignum_t a,istream&is=cin)
{
char buf[MAXX*DIGIT+1],ch ;
int i,j ;
memset((void*)a,0,sizeof(bignum_t));
if(!(is>>buf))return 0 ;
for(a[0]=strlen(buf),i=a[0]/2-1; i>=0; i--)
ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;
for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf); j<a[0]*DIGIT; buf[j++]=‘0‘);
for(i=1; i<=a[0]; i++)
for(a[i]=0,j=0; j<DIGIT; j++)
a[i]=a[i]*10+buf[i*DIGIT-1-j]-‘0‘ ;
for(; !a[a[0]]&&a[0]>1; a[0]--);
return 1 ;
}
void write(const bignum_t a,ostream&os=cout)
{
int i,j ;
for(os<<a[i=a[0]],i--; i; i--)
for(j=DEPTH/10; j; j/=10)
os<<a[i]/j%10 ;
}
int comp(const bignum_t a,const bignum_t b)
{
int i ;
if(a[0]!=b[0])
return a[0]-b[0];
for(i=a[0]; i; i--)
if(a[i]!=b[i])
return a[i]-b[i];
return 0 ;
}
int comp(const bignum_t a,const int b)
{
int c[12]=
{
1
}
;
for(c[1]=b; c[c[0]]>=DEPTH; c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++);
return comp(a,c);
}
int comp(const bignum_t a,const int c,const int d,const bignum_t b)
{
int i,t=0,O=-DEPTH*2 ;
if(b[0]-a[0]<d&&c)
return 1 ;
for(i=b[0]; i>d; i--)
{
t=t*DEPTH+a[i-d]*c-b[i];
if(t>0)return 1 ;
if(t<O)return 0 ;
}
for(i=d; i; i--)
{
t=t*DEPTH-b[i];
if(t>0)return 1 ;
if(t<O)return 0 ;
}
return t>0 ;
}
/************************************************************************/
/* 大数与大数相加 */
/************************************************************************/
void add(bignum_t a,const bignum_t b)
{
int i ;
for(i=1; i<=b[0]; i++)
if((a[i]+=b[i])>=DEPTH)
a[i]-=DEPTH,a[i+1]++;
if(b[0]>=a[0])
a[0]=b[0];
else
for(; a[i]>=DEPTH&&i<a[0]; a[i]-=DEPTH,i++,a[i]++);
a[0]+=(a[a[0]+1]>0);
}
/************************************************************************/
/* 大数与小数相加 */
/************************************************************************/
void add(bignum_t a,const int b)
{
int i=1 ;
for(a[1]+=b; a[i]>=DEPTH&&i<a[0]; a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++);
for(; a[a[0]]>=DEPTH; a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);
}
/************************************************************************/
/* 大数相减(被减数>=减数) */
/************************************************************************/
void sub(bignum_t a,const bignum_t b)
{
int i ;
for(i=1; i<=b[0]; i++)
if((a[i]-=b[i])<0)
a[i+1]--,a[i]+=DEPTH ;
for(; a[i]<0; a[i]+=DEPTH,i++,a[i]--);
for(; !a[a[0]]&&a[0]>1; a[0]--);
}
/************************************************************************/
/* 大数减去小数(被减数>=减数) */
/************************************************************************/
void sub(bignum_t a,const int b)
{
int i=1 ;
for(a[1]-=b; a[i]<0; a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);
for(; !a[a[0]]&&a[0]>1; a[0]--);
}
void sub(bignum_t a,const bignum_t b,const int c,const int d)
{
int i,O=b[0]+d ;
for(i=1+d; i<=O; i++)
if((a[i]-=b[i-d]*c)<0)
a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ;
for(; a[i]<0; a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);
for(; !a[a[0]]&&a[0]>1; a[0]--);
}
/************************************************************************/
/* 大数相乘，读入被乘数a，乘数b，结果保存在c[] */
/************************************************************************/
void mul(bignum_t c,const bignum_t a,const bignum_t b)
{
int i,j ;
memset((void*)c,0,sizeof(bignum_t));
for(c[0]=a[0]+b[0]-1,i=1; i<=a[0]; i++)
for(j=1; j<=b[0]; j++)
if((c[i+j-1]+=a[i]*b[j])>=DEPTH)
c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ;
for(c[0]+=(c[c[0]+1]>0); !c[c[0]]&&c[0]>1; c[0]--);
}
/************************************************************************/
/* 大数乘以小数，读入被乘数a，乘数b，结果保存在被乘数 */
/************************************************************************/
void mul(bignum_t a,const int b)
{
int i ;
for(a[1]*=b,i=2; i<=a[0]; i++)
{
a[i]*=b ;
if(a[i-1]>=DEPTH)
a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ;
}
for(; a[a[0]]>=DEPTH; a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);
for(; !a[a[0]]&&a[0]>1; a[0]--);
}
void mul(bignum_t b,const bignum_t a,const int c,const int d)
{
int i ;
memset((void*)b,0,sizeof(bignum_t));
for(b[0]=a[0]+d,i=d+1; i<=b[0]; i++)
if((b[i]+=a[i-d]*c)>=DEPTH)
b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ;
for(; b[b[0]+1]; b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH);
for(; !b[b[0]]&&b[0]>1; b[0]--);
}
/**************************************************************************/
/* 大数相除,读入被除数a，除数b，结果保存在c[]数组 */
/* 需要comp()函数 */
/**************************************************************************/
void div(bignum_t c,bignum_t a,const bignum_t b)
{
int h,l,m,i ;
memset((void*)c,0,sizeof(bignum_t));
c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ;
for(i=c[0]; i; sub(a,b,c[i]=m,i-1),i--)
for(h=DEPTH-1,l=0,m=(h+l+1)>>1; h>l; m=(h+l+1)>>1)
if(comp(b,m,i-1,a))h=m-1 ;
else l=m ;
for(; !c[c[0]]&&c[0]>1; c[0]--);
c[0]=c[0]>1?c[0]:1 ;
}
void div(bignum_t a,const int b,int&c)
{
int i ;
for(c=0,i=a[0]; i; c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);
for(; !a[a[0]]&&a[0]>1; a[0]--);
}
/************************************************************************/
/* 大数平方根，读入大数a，结果保存在b[]数组里 */
/* 需要comp()函数 */
/************************************************************************/
void sqrt(bignum_t b,bignum_t a)
{
int h,l,m,i ;
memset((void*)b,0,sizeof(bignum_t));
for(i=b[0]=(a[0]+1)>>1; i; sub(a,b,m,i-1),b[i]+=m,i--)
for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1; h>l; b[i]=m=(h+l+1)>>1)
if(comp(b,m,i-1,a))h=m-1 ;
else l=m ;
for(; !b[b[0]]&&b[0]>1; b[0]--);
for(i=1; i<=b[0]; b[i++]>>=1);
}
/************************************************************************/
/* 返回大数的长度 */
/************************************************************************/
int length(const bignum_t a)
{
int t,ret ;
for(ret=(a[0]-1)*DIGIT,t=a[a[0]]; t; t/=10,ret++);
return ret>0?ret:1 ;
}
/************************************************************************/
/* 返回指定位置的数字，从低位开始数到第b位，返回b位上的数 */
/************************************************************************/
int digit(const bignum_t a,const int b)
{
int i,ret ;
for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT; i; ret/=10,i--);
return ret%10 ;
}
/************************************************************************/
/* 返回大数末尾0的个数 */
/************************************************************************/
int zeronum(const bignum_t a)
{
int ret,t ;
for(ret=0; !a[ret+1]; ret++);
for(t=a[ret+1],ret*=DIGIT; !(t%10); t/=10,ret++);
return ret ;
}
void comp(int*a,const int l,const int h,const int d)
{
int i,j,t ;
for(i=l; i<=h; i++)
for(t=i,j=2; t>1; j++)
while(!(t%j))
a[j]+=d,t/=j ;
}
void convert(int*a,const int h,bignum_t b)
{
int i,j,t=1 ;
memset(b,0,sizeof(bignum_t));
for(b[0]=b[1]=1,i=2; i<=h; i++)
if(a[i])
for(j=a[i]; j; t*=i,j--)
if(t*i>DEPTH)
mul(b,t),t=1 ;
mul(b,t);
}
/************************************************************************/
/* 组合数 */
/************************************************************************/
void combination(bignum_t a,int m,int n)
{
int*t=new int[m+1];
memset((void*)t,0,sizeof(int)*(m+1));
comp(t,n+1,m,1);
comp(t,2,m-n,-1);
convert(t,m,a);
delete[]t ;
}
/************************************************************************/
/* 排列数 */
/************************************************************************/
void permutation(bignum_t a,int m,int n)
{
int i,t=1 ;
memset(a,0,sizeof(bignum_t));
a[0]=a[1]=1 ;
for(i=m-n+1; i<=m; t*=i++)
if(t*i>DEPTH)
mul(a,t),t=1 ;
mul(a,t);
}
#define SGN(x) ((x)>0?1:((x)<0?-1:0))
#define ABS(x) ((x)>0?(x):-(x))
int read(bignum_t a,int&sgn,istream&is=cin)
{
char str[MAXX*DIGIT+2],ch,*buf ;
int i,j ;
memset((void*)a,0,sizeof(bignum_t));
if(!(is>>str))return 0 ;
buf=str,sgn=1 ;
if(*buf==‘-‘)sgn=-1,buf++;
for(a[0]=strlen(buf),i=a[0]/2-1; i>=0; i--)
ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;
for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf); j<a[0]*DIGIT; buf[j++]=‘0‘);
for(i=1; i<=a[0]; i++)
for(a[i]=0,j=0; j<DIGIT; j++)
a[i]=a[i]*10+buf[i*DIGIT-1-j]-‘0‘ ;
for(; !a[a[0]]&&a[0]>1; a[0]--);
if(a[0]==1&&!a[1])sgn=0 ;
return 1 ;
}
struct bignum
{
bignum_t num ;
int sgn ;
public :
inline bignum()
{
memset(num,0,sizeof(bignum_t));
num[0]=1 ;
sgn=0 ;
}
inline int operator!()
{
return num[0]==1&&!num[1];
}
inline bignum&operator=(const bignum&a)
{
memcpy(num,a.num,sizeof(bignum_t));
sgn=a.sgn ;
return*this ;
}
inline bignum&operator=(const int a)
{
memset(num,0,sizeof(bignum_t));
num[0]=1 ;
sgn=SGN (a);
add(num,sgn*a);
return*this ;
}
;
inline bignum&operator+=(const bignum&a)
{
if(sgn==a.sgn)add(num,a.num);
else if
(sgn&&a.sgn)
{
int ret=comp(num,a.num);
if(ret>0)sub(num,a.num);
else if(ret<0)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memcpy(num,a.num,sizeof(bignum_t));
sub (num,t);
sgn=a.sgn ;
}
else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
}
else if(!sgn)
memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ;
return*this ;
}
inline bignum&operator+=(const int a)
{
if(sgn*a>0)add(num,ABS(a));
else if(sgn&&a)
{
int ret=comp(num,ABS(a));
if(ret>0)sub(num,ABS(a));
else if(ret<0)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memset(num,0,sizeof(bignum_t));
num[0]=1 ;
add(num,ABS (a));
sgn=-sgn ;
sub(num,t);
}
else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
}
else if
(!sgn)sgn=SGN(a),add(num,ABS(a));
return*this ;
}
inline bignum operator+(const bignum&a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
ret.sgn=sgn ;
ret+=a ;
return ret ;
}
inline bignum operator+(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
ret.sgn=sgn ;
ret+=a ;
return ret ;
}
inline bignum&operator-=(const bignum&a)
{
if(sgn*a.sgn<0)add(num,a.num);
else if
(sgn&&a.sgn)
{
int ret=comp(num,a.num);
if(ret>0)sub(num,a.num);
else if(ret<0)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memcpy(num,a.num,sizeof(bignum_t));
sub(num,t);
sgn=-sgn ;
}
else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
}
else if(!sgn)add (num,a.num),sgn=-a.sgn ;
return*this ;
}
inline bignum&operator-=(const int a)
{
if(sgn*a<0)add(num,ABS(a));
else if(sgn&&a)
{
int ret=comp(num,ABS(a));
if(ret>0)sub(num,ABS(a));
else if(ret<0)
{
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
memset(num,0,sizeof(bignum_t));
num[0]=1 ;
add(num,ABS(a));
sub(num,t);
sgn=-sgn ;
}
else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
}
else if
(!sgn)sgn=-SGN(a),add(num,ABS(a));
return*this ;
}
inline bignum operator-(const bignum&a)
{
bignum ret ;
memcpy(ret.num,num,sizeof(bignum_t));
ret.sgn=sgn ;
ret-=a ;
return ret ;
}
inline bignum operator-(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof(bignum_t));
ret.sgn=sgn ;
ret-=a ;
return ret ;
}
inline bignum&operator*=(const bignum&a)
{
bignum_t t ;
mul(t,num,a.num);
memcpy(num,t,sizeof(bignum_t));
sgn*=a.sgn ;
return*this ;
}
inline bignum&operator*=(const int a)
{
mul(num,ABS(a));
sgn*=SGN(a);
return*this ;
}
inline bignum operator*(const bignum&a)
{
bignum ret ;
mul(ret.num,num,a.num);
ret.sgn=sgn*a.sgn ;
return ret ;
}
inline bignum operator*(const int a)
{
bignum ret ;
memcpy(ret.num,num,sizeof (bignum_t));
mul(ret.num,ABS(a));
ret.sgn=sgn*SGN(a);
return ret ;
}
inline bignum&operator/=(const bignum&a)
{
bignum_t t ;
div(t,num,a.num);
memcpy (num,t,sizeof(bignum_t));
sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ;
return*this ;
}
inline bignum&operator/=(const int a)
{
int t ;
div(num,ABS(a),t);
sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a);
return*this ;
}
inline bignum operator/(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(t,num,sizeof(bignum_t));
div(ret.num,t,a.num);
ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ;
return ret ;
}
inline bignum operator/(const int a)
{
bignum ret ;
int t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(ret.num,ABS(a),t);
ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a);
return ret ;
}
inline bignum&operator%=(const bignum&a)
{
bignum_t t ;
div(t,num,a.num);
if(num[0]==1&&!num[1])sgn=0 ;
return*this ;
}
inline int operator%=(const int a)
{
int t ;
div(num,ABS(a),t);
memset(num,0,sizeof (bignum_t));
num[0]=1 ;
add(num,t);
return t ;
}
inline bignum operator%(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(t,ret.num,a.num);
ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ;
return ret ;
}
inline int operator%(const int a)
{
bignum ret ;
int t ;
memcpy(ret.num,num,sizeof(bignum_t));
div(ret.num,ABS(a),t);
memset(ret.num,0,sizeof(bignum_t));
ret.num[0]=1 ;
add(ret.num,t);
return t ;
}
inline bignum&operator++()
{
*this+=1 ;
return*this ;
}
inline bignum&operator--()
{
*this-=1 ;
return*this ;
}
;
inline int operator>(const bignum&a)
{
return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0);
}
inline int operator>(const int a)
{
return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0);
}
inline int operator>=(const bignum&a)
{
return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0);
}
inline int operator>=(const int a)
{
return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0);
}
inline int operator<(const bignum&a)
{
return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0);
}
inline int operator<(const int a)
{
return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0);
}
inline int operator<=(const bignum&a)
{
return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0);
}
inline int operator<=(const int a)
{
return sgn<0?(a<0?comp(num,-a)>=0:1):
(sgn>0?(a>0?comp(num,a)<=0:0):a>=0);
}
inline int operator==(const bignum&a)
{
return(sgn==a.sgn)?!comp(num,a.num):0 ;
}
inline int operator==(const int a)
{
return(sgn*a>=0)?!comp(num,ABS(a)):0 ;
}
inline int operator!=(const bignum&a)
{
return(sgn==a.sgn)?comp(num,a.num):1 ;
}
inline int operator!=(const int a)
{
return(sgn*a>=0)?comp(num,ABS(a)):1 ;
}
inline int operator[](const int a)
{
return digit(num,a);
}
friend inline istream&operator>>(istream&is,bignum&a)
{
read(a.num,a.sgn,is);
return is ;
}
friend inline ostream&operator<<(ostream&os,const bignum&a)
{
if(a.sgn<0)
os<<‘-‘ ;
write(a.num,os);
return os ;
}
friend inline bignum sqrt(const bignum&a)
{
bignum ret ;
bignum_t t ;
memcpy(t,a.num,sizeof(bignum_t));
sqrt(ret.num,t);
ret.sgn=ret.num[0]!=1||ret.num[1];
return ret ;
}
friend inline bignum sqrt(const bignum&a,bignum&b)
{
bignum ret ;
memcpy(b.num,a.num,sizeof(bignum_t));
sqrt(ret.num,b.num);
ret.sgn=ret.num[0]!=1||ret.num[1];
b.sgn=b.num[0]!=1||ret.num[1];
return ret ;
}
inline int length()
{
return :: length(num);
}
inline int zeronum()
{
return :: zeronum(num);
}
inline bignum C(const int m,const int n)
{
combination(num,m,n);
sgn=1 ;
return*this ;
}
inline bignum P(const int m,const int n)
{
permutation(num,m,n);
sgn=1 ;
return*this ;
}
};
/**+++++++++++++++++++分界线++++++++++++++++++++++**/
bignum ans[105];
bignum c[105][105];
void Init()
{
for(int i=1; i<105; i++)
{
c[0][i] = 1;
c[i][0] = 1;
}
for(int i=1; i<105; i++)
{
for(int j=1; j<=i; j++)
{
c[i][j] = c[i-1][j-1]+c[i-1][j];
}
}
}
void Init2()
{
ans[0] = 1, ans[1] = 0, ans[2] = 1;
bignum ret;
ret = 1;
int last = 0;
for(int i=3; i<102; i++)
{
int x = i*(i-1)/2;
bignum tmp;
tmp = 0;
for(int j=1; j<=i; j++)
tmp += c[i][j]*ans[i-j];
for(int j = last; j<x; j++)
ret = ret*2;
last = x;
ans[i] = ret - tmp;
}
}
int main()
{
Init();
Init2();
short n;
while(cin>>n)
{
cout<<ans[n]<<endl;
}
return 0;
}
时间： 2025-01-16 18:11:03