hdu_2838_Cow Sorting(树状数组求逆序对)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2838

题意：给你一串数，让你排序，只能交换相邻的数，每次交换花费交换的两个树的和，问最小交换的价值

题解：实质就是求逆序对

 #include<cstdio>
 #define F(i,a,b) for(int i=a;i<=b;i++)
 typedef long long LL;
 const int N=1e5+;
 int num[N],n,x,cnt;
 LL sum[N],ans;    

 inline void add(int x,int k){for(;x<=;x+=x&-x)num[x]+=,sum[x]+=k;}
 inline int asknum(int x){int an=;while(x>)an+=num[x],x-=x&-x;return an;}
 inline LL asksum(int x){LL an=;while(x>)an+=sum[x],x-=x&-x;return an;}

 int main(){
     while(~scanf("%d",&n)){
         F(i,,N-)num[i]=,sum[i]=;ans=;
         F(i,,n){
             scanf("%d",&x),add(x,x),cnt=i-asknum(x);
             if(cnt)ans+=(LL)cnt*x+asksum()-asksum(x);
         }
         printf("%lld\n",ans);
     }
     return ;
 }