HDU 5775 （Bubble Sort 逆序数）

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题目链接
意思是告诉你一个冒泡排序，在整个排序的过程中问你每个数出现最左与最右的位置差

其实就是求每个数后面有几个比他小的，因为后面小的排序会移到他前面，用树状数组写一下就可以了

#include<cstdio>#include<algorithm>#include<iostream>#include<vector>#include<queue>#include<cstring>#include<bits/stdc++.h>#include<map>#include<string>#include<cstdlib>using namespace std;#define cl(a,b) memset(a,b,sizeof(a))#define LL long long#define pb push_back#define gcd __gcd#define For(i,j,k) for(int i=(j);i<k;i++)#define lowbit(i) (i&(-i))#define _(x) printf("%d\n",x)const int maxn = 1e6+200;const int inf  = 1 << 28;struct BIT{    LL c[maxn],sz;    void init(int s){        sz=s;for(int i=0;i<=sz;c[i]=0,i++);    }    void updata(int x,int val){        while(x<=sz){            c[x]+=val;            x+=lowbit(x);        }    }    LL sum(int x){        LL sum=0;        while(x>0){            sum+=c[x];            x-=lowbit(x);        }        return sum;    }}bit;int a[maxn];int ans[maxn];int main(){    int T;scanf("%d",&T);    int cas = 1;    while(T--){        int n;scanf("%d",&n);        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        printf("Case #%d:",cas++);        bit.init(n);        for(int i=n;i>=1;i--){            bit.updata(a[i],1);            int x = bit.sum(a[i])-1;            int right = max(a[i],i+x);            int left = min(a[i],i);            int y=abs(left-right);            ans[a[i]]=y;        }        for(int i=1;i<=n;i++)printf(" %d",ans[i]);puts("");    }    return 0;}

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