1009. Triple Inversions (35)

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IDEA

1.使用树状数组求逆序数的问题，参考 http://wenku.baidu.com/view/8251fc0516fc700abb68fc76.html?re=view 点击打开链接

2.以中间数为基准，算出左边比其大的数个数，右边比其小的数个数，二者相乘得出结果

3.结果用long long表示

CODE

#include<iostream>#include<vector> #include<queue>#include<algorithm>#include<fstream>using namespace std;#define MAX 100005int n;int a[MAX],c[MAX]={0};long long l[MAX],r[MAX];int lowbit(int x){return x&-x;}void add(int x){while(x<=n){c[x]++;x+=lowbit(x);}}int sum(int x){int s=0;while(x>0){s+=c[x];x-=lowbit(x);}return s;}int main(){#ifndef ONLINE_JUDGEfreopen("input.txt","r",stdin);#endifcin>>n;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=n;i++){add(a[i]);l[i]=i-sum(a[i]);//i为当前已经插入数的个数，sum(a[i])为比a[i]小的数的个数，i-sum(a[i])即为比 比a[i]大的数的个数，即逆序个数 }for(int i=1;i<=n;i++){c[i]=0;}//for(int i=1;i<=n;i++){//add(a[i]);//r[i]=a[i]-sum(a[i]);//}for(int i=n;i>=1;i--){r[i]=sum(a[i]);add(a[i]);}long long count=0;for(int i=1;i<=n;i++){count+=l[i]*r[i];}cout<<count<<endl;#ifndef ONLINE_JUDGEfclose(stdin);#endifreturn 0;}

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