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+7;
int num[N],n,x,cnt;
LL sum[N],ans;    

inline void add(int x,int k){for(;x<=100000;x+=x&-x)num[x]+=1,sum[x]+=k;}
inline int asknum(int x){int an=0;while(x>0)an+=num[x],x-=x&-x;return an;}
inline LL asksum(int x){LL an=0;while(x>0)an+=sum[x],x-=x&-x;return an;}

int main(){
    while(~scanf("%d",&n)){
        F(i,1,N-1)num[i]=0,sum[i]=0;ans=0;
        F(i,1,n){
            scanf("%d",&x),add(x,x),cnt=i-asknum(x);
            if(cnt)ans+=(LL)cnt*x+asksum(100000)-asksum(x);
        }
        printf("%lld
",ans);
    }
    return 0;
}