Rotate

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

题目大意：求顺序数对数，但是有转移操作

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2539    Accepted Submission(s): 575

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

  

   5
1 2 3 4 5
3
Q
R 1 3
Q
  

 

Sample Output

  

   10
8
  

 

Author

yifenfei

 

Source

ZJFC 2009-3 Programming Contest

 

这个代码可能过不去，看人品。998ms过得，未来优化

#include <cstdio>
#include <cstring>
#define maxn 3000010
int n;
int a[maxn],c[maxn];
int bit(int t){
    return t&(-t);
}
void update(int pos){
    while(pos<=10000){
        c[pos]++;
        pos+=bit(pos);
    }
}
__int64 getsum(int pos){
    __int64 ans=0;
    while(pos>0){
        ans+=c[pos];
        pos-=bit(pos);
    }
    return ans;
}
int main()
{
    __int64 ans;
    while(scanf("%d",&n)!=EOF){
        memset(c,0,sizeof(c));
        ans=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            ans+=getsum(a[i]-1);//避免有重复的a[i]
            update(a[i]);
        }
        int m;
        scanf("%d",&m);
        while(m--){
            char ch[3];
            scanf("%s",ch);
            if(ch[0]=='Q'){
                printf("%I64d\n",ans);
            }
            else {
                int s,e;
                scanf("%d%d",&s,&e);
                int cnt=a[s];
                for(int i=s;i<e;i++){
                    a[i]=a[i+1];
                    if(a[i]>cnt)ans--;
                    else if(a[i]<cnt)ans++;
                }
                a[e]=cnt;
            }
        }
    }
    return 0;
}