题目大意
给一个排列，删除\(m\)个元素，每删一个前，输出逆序对个数。
\(\text{Solution}\)
一道较为简单的\(CDQ\)分治，考虑每删一个数，后逆序对数个数的变化，即上次的\(ans\)减删除数前比他大的，后比他小的。
那么我们可以静态的求这个数，给每一个数加一个消失时间\(T_i\)，当\(T_i > T_j\)时，\(j\)才能对\(i\)产生贡献。
码代码时很轻松，但忘了开\(long\) \(long\)，调试了好久，所以······
\(\text{Code}\)
#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std;
const int N = 1e5 + 5;
LL b[N],c[N],inc[N],ans = 0,f[N];
int n,m;
struct nd{int x,y,z;}a[N],na[N];
int lowbit(int x){return x & -x;}
void add(int x,int v){while (x <= n) f[x] += v,x += lowbit(x);}
LL query(int x)
{
LL res = 0;
while (x) res += f[x],x -= lowbit(x);
return res;
}
bool cmp(nd x,nd y){return x.x > y.x;}
void solve(int l,int r,int fl)
{
if (l == r) return;
int mid = l + r >> 1;
solve(l,mid,fl),solve(mid + 1,r,fl);
int i = l,j = mid + 1,cnt = l;
while (i <= mid && j <= r)
{
if (a[i].z > a[j].z) add(a[i].y,1),na[cnt++] = a[i],i++;
else
{
if (fl) inc[a[j].y] += query(a[j].y);
else inc[a[j].y] += query(n) - query(a[j].y);
na[cnt++] = a[j],j++;
}
}
while (i <= mid) add(a[i].y,1),na[cnt++] = a[i],i++;
while (j <= r)
{
if (fl) inc[a[j].y] += query(a[j].y);
else inc[a[j].y] += query(n) - query(a[j].y);
na[cnt++] = a[j],j++;
}
for (int k = l; k <= mid; k++) add(a[k].y,-1);
for (int k = l; k <= r; k++) a[k] = na[k];
}
int main()
{
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; i++)
a[i].x = i,scanf("%lld",&a[i].y),ans += query(n) - query(a[i].y),add(a[i].y,1);
for (int i = 1; i <= n; i++) add(a[i].y,-1),b[i] = n;
for (int i = 1; i <= m; i++) scanf("%lld",&c[i]),b[c[i]] = i;
for (int i = 1; i <= n; i++) a[i].z = b[a[i].y];
solve(1,n,0),sort(a + 1,a + 1 + n,cmp),solve(1,n,1);
for (int i = 1; i <= m; i++) printf("%lld\n",ans),ans -= inc[c[i]],inc[c[i]] = 0;
}