SGU 180 Inversions【树状数组】

题意：求逆序数

和POJ那题求逆序数的一样，不过这题离散化之后，要去一下重，然后要开到long long

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<stack>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;

 typedef long long LL;
 const int INF = (<<)-;
 const int mod=;
 const int maxn=;

 int a[maxn];
 int c[maxn];//树状数组

 struct node{
     int x;
     int id;
 } p[maxn];

 int cmp(node n1,node n2){
     return n1.x < n2.x;
 }

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

 __int64 sum(int x){
     __int64 ret=;
     while(x > ){
         ret+=c[x];x-=lowbit(x);
     }
     return ret;
 }

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

 int main(){
     int n;
     scanf("%d",&n);
     for(int i=;i<=n;i++) scanf("%d",&p[i].x),p[i].id=i;
     sort(p+,p+n+,cmp);

     for(int i=,j=;i<=n;i++){
         if(i== || p[i].x != p[i-].x) j++;
         a[p[i].id] = j;
     }

     memset(c,,sizeof(c));
     __int64 ans=;
     for(int i=;i <=n;i++ ) {
         add(a[i],);
         ans += i - sum(a[i]);
     }
     printf("%I64d\n",ans);
     return ;
 }