HDU2688 树状数组（逆序数）

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3709    Accepted Submission(s): 711

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

 

Author

yifenfei

 

Source

ZJFC 2009-3 Programming Contest

 题意：

共n个数，m次操作，Q表示询问n个数有多少顺序数，R a,b表示区间[a,b]中的数转动一下（a+1~b每个数向前移动，a位置的数到b位置）.

代码：

//如果每询问一次就求一次顺序数很费时间，先求出最初的顺序数ans，，每转动一次时
//如果a+1~b位置的数大于a位置的数,ans--,如果小于a，ans++。hduoj时而TLE时而AC.
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long A[10004];
int f[3000006];
int Lowbit(int id) {return id&(-id);}
void Add(int id,int c)
{
    while(id<=10000){
        A[id]+=c;
        id+=Lowbit(id);
    }
}
long long query(int id)
{
    long long s=0;
    while(id>0){
        s+=A[id];
        id-=Lowbit(id);
    }
    return s;
}
int main()
{
    int n,m,a,b;
    char ch[3];
    while(scanf("%d",&n)==1){
        memset(A,0,sizeof(A));
        long long ans=0;
        for(int i=0;i<n;i++){
            scanf("%d",&f[i]);
            Add(f[i],1);
            ans+=query(f[i]-1);
        }
        scanf("%d",&m);
        while(m--){
            scanf("%s",ch);
            if(ch[0]=='Q') printf("%I64d
",ans);
            else{
                scanf("%d%d",&a,&b);
                int v=f[a];
                for(int i=a;i<b;i++){
                    f[i]=f[i+1];
                    if(f[i]>v) ans--;
                    else if(f[i]<v) ans++;
                }
                f[b]=v;
            }
        }
    }
    return 0;
}