HDU-2688-Rotate（树状数组）

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. 

InputThe 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. 
OutputOutput just according to said.Sample Input

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

Sample Output

10
8

题解

这道题我们可以用树状数组求出原数组的答案

因为abs(E-S)<=1000，m<10000，所以我们可以枚举每次的区间，因为翻转每次是S+1~E往前移一位，第S个到E

所以我们先把第S位存下来，每次和后面的判断一下大小就可以了

当输入是Q的时候就直接输出就可以了

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath> 
#define ll long long
#define M 10005
#define N 3000005
using namespace std;
int n,m,x,y;
ll ans;
int a[N];
ll tr[M];
char s[10];
int lowbit(int x){ return x&(-x); }
void add(int x){
    while (x<=M){
        tr[x]++;
        x+=lowbit(x);
    }
}
int query(int x){
    int s=0;
    while (x>0){
        s+=tr[x];
        x-=lowbit(x);
    }
    return s;
}
int main(){
    while (~scanf("%d",&n)){
        memset(tr,0,sizeof(tr)); ans=0;
        for (int i=0;i<n;i++){
            scanf("%d",&a[i]);
            add(a[i]);
            ans+=query(a[i]-1);
        }
        scanf("%d",&m);
        for (int i=1;i<=m;i++){
            scanf("%s",s);
            if (s[0]=='Q') printf("%lld
",ans);
            else{
                scanf("%d%d",&x,&y);
                int s=a[x];
                for (int i=x;i<y;i++){
                    a[i]=a[i+1];
                    if (s<a[i]) ans--; else
                    if (s>a[i]) ans++;
                }
                a[y]=s;
            }
        }
    }
    return 0;
}