1003（树状数组+逆序对）

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

题目大概：

求逆序对数，并且要让一个区间的数循环一遍，找出最大的逆序对数。

思路：
先按照模板把逆序对数求出来，然后把区间的数依次循环，求出每个数字位置改变后的新逆序对数。然后选出最大的。

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[3000000];int c[10001];int lowbit(int x){    return x&(-x);}void add(int x,int v){    while(x<=10001)    {        c[x]+=v;        x+=lowbit(x);    }}long long sum(int x){    long long sum=0;    while(x>0)    {        sum+=c[x];        x-=lowbit(x);    }    return sum;}int main(){    int n,m;    while(~scanf("%d",&n))    {    long long ans=0;    memset(c,0,sizeof(c));    for(int i=0;i<n;i++)    {        scanf("%d",&a[i]);        ans+=sum(a[i]-1);        add(a[i],1);    }    scanf("%d",&m);    char fi[2];    int q,w,vv;    while(m--)    {        scanf("%s",fi);        if(fi[0]=='Q')        {            printf("%I64d\n",ans);        }        else        {            scanf("%d%d",&q,&w);            vv=a[q];            for(int i=q;i<w;i++)            {                a[i]=a[i+1];                if(vv>a[i])ans++;                if(vv<a[i])ans--;            }            a[w]=vv;        }    }    }    return 0;}