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

59105431230

Sample Output

60

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;}