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

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

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath> 
 5 #define ll long long
 6 #define M 10005
 7 #define N 3000005
 8 using namespace std;
 9 int n,m,x,y;
10 ll ans;
11 int a[N];
12 ll tr[M];
13 char s[10];
14 int lowbit(int x){ return x&(-x); }
15 void add(int x){
16     while (x<=M){
17         tr[x]++;
18         x+=lowbit(x);
19     }
20 }
21 int query(int x){
22     int s=0;
23     while (x>0){
24         s+=tr[x];
25         x-=lowbit(x);
26     }
27     return s;
28 }
29 int main(){
30     while (~scanf("%d",&n)){
31         memset(tr,0,sizeof(tr)); ans=0;
32         for (int i=0;i<n;i++){
33             scanf("%d",&a[i]);
34             add(a[i]);
35             ans+=query(a[i]-1);
36         }
37         scanf("%d",&m);
38         for (int i=1;i<=m;i++){
39             scanf("%s",s);
40             if (s[0]=='Q') printf("%lld
",ans);
41             else{
42                 scanf("%d%d",&x,&y);
43                 int s=a[x];
44                 for (int i=x;i<y;i++){
45                     a[i]=a[i+1];
46                     if (s<a[i]) ans--; else
47                     if (s>a[i]) ans++;
48                 }
49                 a[y]=s;
50             }
51         }
52     }
53     return 0;
54 }

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