树状数组_Rotate HDU2688

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

51 2 3 4 53QR 1 3Q

 

Sample Output

108
题意：给定一个区间，求满足(ai<aj && i<j)的序对的个数.遇到Q输出答案，遇到R对一个子区间进行rotate.比如，子区间a,b,c进行rotate后变成b,c,a.值得注意的是给定R 1 3,实际子区间是[2,4].
分析：用树状数组能很好的求出原序列中正序对的个数ans，然后只需对子区间进行rotate并更新ans值.
这题时间卡得有点无语，963ms过的.并且有时会过不了.
AC代码：
#include <cstdio>#include <cstring>using namespace std;#define lowbit(x) (x&-x)#define maxn 3000030typedef long long ll;ll n,c[10005],a[maxn];ll sum(ll x){    ll res=0;    while(x>0)    {        res+=c[x];        x-=lowbit(x);    }    return res;}void add(ll x,ll d){    while(x<=10005)    {        c[x]+=d;        x+=lowbit(x);    }}int main(){    while(scanf("%lld",&n)!=EOF)    {        for(ll i=1;i<=n;i++)            scanf("%lld",&a[i]);        memset(c,0,sizeof(c));        ll ans=0;        for(ll i=1;i<=n;i++)        {            add(a[i],1);            ans+=sum(a[i]-1);        }        ll t;        scanf("%lld",&t);        while(t--)        {            char ch;            getchar();            scanf("%c",&ch);            if(ch=='R')            {                ll l,r;                scanf("%lld%lld",&l,&r);                l++,r++;                ll tmp=a[l];                for(ll i=l+1;i<=r;i++)                {                    if(tmp<a[i]) ans--;                    else if(tmp>a[i]) ans++;                    a[i-1]=a[i];                }                a[r]=tmp;            }            else if(ch=='Q')            {            printf("%lld\n",ans);            }        }    }    return 0;}