HDOJ 2688 HDU 2688 Rotate ACM 2688 IN HDU

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋     

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=2688

题目描述: 

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 250    Accepted Submission(s): 68

Problem Description

Recently yifenfei face such a problem that give you millions of positive integers,tell how many pairs i and j that satisfy F[i] smaller than F[j] strictly when i is smaller than j strictly. i and j is the serial number in the interger sequence. Of course, the problem is not over, the initial interger sequence will change all the time. Changing format is like this [S E] (abs(E-S)<=1000) that mean between the S and E of the sequece will Rotate one times.
For example initial sequence is 1 2 3 4 5.
If changing format is [1 3], than the sequence will be 1 3 4 2 5 because the first sequence is base from 0.

 

Input

The input contains multiple test cases.
Each case first given a integer n standing the length of integer sequence (2<=n<=3000000) 
Second a line with n integers standing F[i](0<F[i]<=10000)
Third a line with one integer m (m < 10000)
Than m lines quiry, first give the type of quiry. A character C, if C is ‘R’ than give the changing format, if C equal to ‘Q’, just put the numbers of satisfy pairs.

 

Output

Output just according to said.

 

Sample Input

5 1 2 3 4 5 3 Q R 1 3 Q

 

Sample Output

10 8

 

 题目分析:

如果是暴力 , 没一次更新都要重新计算的话,  时间上的开销会非常大.

对数据进行分析,可以看到, 当旋转的时候, 除了 第一个数, 其他的数的对数的个数值都与这个数有关,   因此, 只要开始

先把总的对数 sum 算出来, 再 根据旋转时 每个数跟第一个数的大小 比较 ,对sum 进行更新就可以了.

 

代码如下:

代码

/*
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
           http://www.cnblog.com/MiYu
Author By : MiYu
Test      : 1
Program   : 2688
*/

#include  < iostream >
#include  < algorithm >
using   namespace  std;
const   int  MAX  =   10000 ;
int  nCount  =   0 , N, M, x , y;
int  com[MAX  +   1 ], num[ 300   *  MAX  +   1 ]; 
long   long  sum  =   0 ; 
char  ask[ 5 ];
inline  int  low (  int  x ) {
        return  x  &  (  - x );
}
void  modify (  int  x,  int  val ){      //  修改 
      while  ( x  <=  MAX ){
            com[x]  +=  val; x  +=  low ( x );
     }     
}
int  quy (  int  x ){                   //  查询 
     int  sum  =   0 ;
     while  ( x  >   0  ){
           sum  +=  com[x]; x  ^=  low ( x );
    }
     return  sum ;
}
inline  bool  scan_d( int   & num)       //  输入 
{
         char   in ; bool  IsN = false ;
         in = getchar();
         if ( in == EOF)  return   false ;
         while ( in != ' - ' && ( in < ' 0 ' || in > ' 9 ' ))  in = getchar();
         if ( in == ' - ' ){ IsN = true ;num = 0 ;}
         else  num = in - ' 0 ' ;
         while ( in = getchar(), in >= ' 0 ' && in <= ' 9 ' ){
                num *= 10 ,num += in - ' 0 ' ;
        }
         if (IsN) num =- num;
         return   true ;
}
int  main ()
{
     while  ( scan_d ( N ) ){
           memset ( com,  0 ,  sizeof  ( com ) ); sum  =   0 ;
            for  (  int  i  =   0 ; i  <  N;  ++  i ){
                scan_d ( num[i] ); modify ( num[i],  1  ); 
                sum  +=  quy ( num[i]  -   1  );
           }
           scan_d ( M );
            while  ( M  --  ){
                  scanf (  " %s " ,ask );   int  temp; 
                   switch  ( ask[ 0 ] ){
                           case   ' Q '  : cout  <<  sum  <<  endl;  break ;
                           case   ' R '  : scan_d ( x ), scan_d ( y ); temp  =  num[x];
                                      while  ( x  <  y ) { num[x]  =  num[x + 1 ]; 
                                             num[x]  >  temp  ?  sum  --  : num[x]  ==  temp  ? : sum ++  ; x  ++ ; 
                                     } 
                                     num[y]  =  temp;  break ;                             
                  }
           }
    }
     return   0 ;
}

 

 

 

 

 

转载于:https://www.cnblogs.com/MiYu/archive/2010/08/30/1812690.html