1020（树状数组，逆序对+离散化）

Problem Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence

9 1 0 5 4 ,

Ultra-QuickSort produces the output

0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

题目大概+思路：

逆序对数量的模板题。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;


int c[500001];
int b[500002];
int lowbit(int x)
{
    return x&(-x);
}

void add(int x,int v)
{
    while(x<=500001)
    {
        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;
}

struct poin
{
    int v,id;
}a[500005];

int cmp(const poin a,const poin b)
{
    if(a.v<b.v)return 1;
    else return 0;
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {

    memset(c,0,sizeof(c));
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].v);
        a[i].id=i;

    }
    sort(a+1,a+n+1,cmp);

    b[a[1].id]=1;
    int o=2;
    for(int i=2;i<=n;i++)
    {
        if(a[i].v==a[i-1].v)b[a[i].id]=b[a[i-1].id];
        else b[a[i].id]=o++;
    }


   long long su=0;
    for(int j=1;j<=n;j++)
    {
        add(b[j],1);
        su+=j-sum(b[j]);

    }
     printf("%I64d\n",su);
    }

    return 0;
}