poj2299 树状数组入门题

题意：利用树状数组求逆序数；

思路：因为输入范围较大，先离散化一下，得到的数组a记录了原来数组的大小关系；然后按下标顺序执行add(a[i],1)，这样sum(a[i])得到的就是小于等于a[i]的个数，i-sum(a[i])即为a[i]前面比a[i]大的数的个数 //外循环n次并累加i-sum(a[i])得到逆序数

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

struct node{
   int v,order;
}a[500050];

bool cmp(node a, node b){
   return a.v < b.v;
}

int aa[500050],c[500050];
int n;

int lowbit(int x)
{
    return x & (-x);
}

int sum(int x){
    int ret = 0;
    while(x > 0)
    {
        ret += c[x]; x -= lowbit(x);
    }
    return ret;
}

void add(int x, int d){
    while(x <= n){
        c[x] += d; x += lowbit(x);
    }
}

int main()
{
    while(scanf("%d",&n) == 1 && n)
    {
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d",&a[i].v);
            a[i].order = i;
        }
        sort(a+1,a+n+1,cmp);
        for(int i = 1; i <= n; ++i)
        {
            aa[a[i].order] = i;
        }
        memset(c,0,sizeof(c));
        LL ans = 0;
        for(int i = 1; i <= n; ++i)
        {
            add(aa[i],1);
            ans += i-sum(aa[i]);
        }
        printf("%lld
",ans);
    }
    return 0;
}