hdu Minimum Inversion Number—樹狀數組求正序數與逆序數

/*數組初始化為0後插入1統計的即為正序數C(n,2)-正序數為逆序數又題意求的為最小的逆序數且變換規則為a1, a2, ..., an-1, an (where m = 0 - the initial seqence)a2, a3, ..., an, a1 (where m = 1)a3, a4, ..., an, a1, a2 (where m = 2)...an, a1, a2, ..., an-1 (where m = n-1)a1後有an-a1個比a1大的，（a1-1）個比a1小的(-1指本身),然後求解ans+=(n-a1)-(a1-1)*/#include<stdio.h>#include<string.h>#define MAXN 5050int c[MAXN];int n;int lowbit(int x){return x&(-x);}void update(int i,int val){while(i<=n){c[i]+=val;i+=lowbit(i);}}int query(int i){int s=0;while(i>0){s+=c[i];i-=lowbit(i);}return s;}int a[MAXN];int main(){int i;while(scanf("%d",&n)!=EOF){int ans=0;memset(c,0,sizeof(c));for(i=1;i<=n;i++){scanf("%d",&a[i]);a[i]++;ans+=query(a[i]);update(a[i],1);}int min=n*(n-1)/2-ans;ans=min;//這個地方之前沒注意到一直WAfor(i=1;i<=n;i++){ans+=n-a[i]-(a[i]-1);if(ans<min)min=ans;}printf("%d\n",min);}return 0;}