hdu 2688_Rotate(树状数组，求顺序对数）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2688

题目大意：求一序列的顺序数对有几个，对应着不同的操作，有查询和更改；

题目思路：用A[i]存序列，注意从0开始存的，然后C[i]存小于和等于i的数有几个

题目：

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1463    Accepted Submission(s): 380

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

 

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int size=10005;int A[3000000],C[size];int lowbit(int x){    //return x&(x^(x-1));    return x&-x;}__int64 sum(int end){    __int64 ans=0;    while(end>0)    {        ans+=C[end];        end-=lowbit(end);    }    return ans;}void updata(int pos,int num){    while(pos<size)    {        C[pos]+=num;        pos+=lowbit(pos);    }}int main(){    int N,M;    while(scanf("%d",&N)!=EOF)    {        __int64 ans=0;        memset(C,0,sizeof(C));        for(int i=0; i<N; i++)        {            scanf("%d",&A[i]);            ans+=sum(A[i]-1);            updata(A[i],1);        }        scanf("%d",&M);        char op[10];        int ss,ee;        while(M--)        {            scanf("%s",op);            if(op[0]=='Q')            {                printf("%I64d\n",ans);            }            else            {                scanf("%d%d",&ss,&ee);                int v=A[ss];                for(int i=ss; i<ee; i++)                {                    A[i]=A[i+1];                    if(v<A[i])                    {                        ans--;                    }                    if(v>A[i])                    {                        ans++;                    }                }                A[ee]=v;            }        }    }    return 0;}