题意:
给一个特大数字长度<=100000,现在依次将最后一位数字放置在第一位,求这些数字中比初始数字小的有多少个,相等的有多少个,大的有多少个(数字不可以重复)。
思路:
对原字符串进行倍增后作为母串S,原字符串作为子串T,进行扩展KMP处理。对于Ex[i]==len的位置说明以此处为初始的字符串与原串相等,当Ex[i]<len时我们只需比较S[i+Ex[i]]与T[Ex[i]]即可。注意数字不可以重复,所以我们需要计算一下循环节!
C++代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
const int maxm = 200010;
int Next[maxn],Ex[maxm];
void GetKmpNext( char *s , int *Next )
{
Next[0] = -1;
int len = strlen(s);
for ( int i=0,j=-1 ; i<len ; i++ )
{
while( j!=-1&&s[i]!=s[j] )
j = Next[j];
Next[i+1] = ++j;
}
}
void GetNext( char *s , int *Next )
{
int c = 0,len = strlen(s);
Next[0] = len;
while( s[c]==s[c+1]&&c+1<len ) c++;
Next[1] = c;
int po = 1;
for ( int i=2 ; i<len ; i++ )
{
if ( Next[i-po]+i<Next[po]+po )
Next[i] = Next[i-po];
else
{
int t = Next[po]+po-i;
if ( t<0 ) t = 0;
while( i+t<len&&s[t]==s[i+t] ) t++;
Next[i] = t;
po = i;
}
}
}
void ExKmp( char *s1 , char *s2 , int *Next , int *Ex )
{
int c = 0,len = strlen(s1),l2 = strlen(s2);
GetNext( s2 , Next );
while( s1[c]==s2[c]&&c<len&&c<l2 ) c++;
Ex[0] = c;
int po = 0;
for ( int i=1 ; i<len ; i++ )
{
if ( Next[i-po]+i<Ex[po]+po )
Ex[i] = Next[i-po];
else
{
int t = Ex[po]+po-i;
if ( t<0 ) t = 0;
while ( i+t<len&&t<l2&&s1[i+t]==s2[t] ) t++;
Ex[i] = t;
po = i;
}
}
}
char SS[maxm],S[maxn];
int main()
{
int Cas; scanf ( "%d" , &Cas );
for ( int cas=1 ; cas<=Cas ; cas++ )
{
scanf ( "%s" , S );
GetKmpNext( S , Next );
int len = strlen(S);
int d = (len%(len-Next[len])==0)?len/(len-Next[len]):1;
strcpy( SS , S );
for ( int i=len ; i<len+len ; i++ )
SS[i] = S[i-len];
SS[len+len] = '\0';
ExKmp( SS , S , Next , Ex );
int a = 0,b = 0,c = 0;
for ( int i=0 ; i<len ; i++ )
{
if ( Ex[i]==len ) b++;
else if ( SS[i+Ex[i]]<S[Ex[i]] ) a++;
else if ( SS[i+Ex[i]]>S[Ex[i]] ) c++;
}
printf ( "Case %d: %d %d %d\n" , cas , a/d , b/d , c/d );
}
return 0;
}