HDU2688-Rotate

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[];
 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=;
     while (x>){
         s+=tr[x];
         x-=lowbit(x);
     }
     return s;
 }
 int main(){
     while (~scanf("%d",&n)){
         memset(tr,,sizeof(tr)); ans=;
         for (int i=;i<n;i++){
             scanf("%d",&a[i]);
             add(a[i]);
             ans+=query(a[i]-);
         }
         scanf("%d",&m);
         for (int i=;i<=m;i++){
             scanf("%s",s);
             if (s[]=='Q') printf("%lld\n",ans);
             else{
                 scanf("%d%d",&x,&y);
                 int s=a[x];
                 for (int i=x;i<y;i++){
                     a[i]=a[i+];
                     if (s<a[i]) ans--; else
                     if (s>a[i]) ans++;
                 }
                 a[y]=s;
             }
         }
     }
     return ;
 }