题目里面的将第一张放到牌堆最下面看起来很玄乎,其实就是摸牌的时候循环摸牌
我们的目标是在摸完第i张牌后,迅速判断要摸到第i+1张牌要再摸几次
于是我们可以用线段树来维护一个区间内还有多少牌没有被扔出去,这样每次做一个区间查询,扔牌时做一个单点修改就好
这题还有一个难点是如何确定牌的大小次序
对于序号不同的牌,当然是序号小的靠前
对于序号相同的牌,要看上一种牌的最后一张在哪里,然后按照
1.在它后面的一定比在它前面的小
2.都在它后面的离它越近越小
3.都在它前面的离它越远越小
进行排序
注意:牌上的序号是<=1e5,不是n
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <utility>
#include <map>
#include <stack>
#include <set>
#include <vector>
#include <queue>
#include <deque>
#include <bitset>
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define LL long long
#define Pair pair<int,int>
#define LOWBIT(x) x & (-x)
using namespace std;
const int zero_stand=1500;
const int MOD=1e9+7;
const int INF=0x7ffffff;
const int magic=348;
vector<int> v[100048];
int stand;
struct node
{
int left,right;
int sum;
}tree[300048];
void build(int cur,int left,int right)
{
tree[cur].left=left;tree[cur].right=right;tree[cur].sum=right-left+1;
if (left!=right)
{
int mid=(left+right)>>1;
build(cur*2,left,mid);
build(cur*2+1,mid+1,right);
}
}
void update(int cur,int pos)
{
if (tree[cur].left==tree[cur].right)
{
tree[cur].sum=0;
return;
}
int mid=(tree[cur].left+tree[cur].right)>>1;
if (pos<=mid) update(cur*2,pos); else update(cur*2+1,pos);
tree[cur].sum=tree[cur*2].sum+tree[cur*2+1].sum;
}
int query(int cur,int left,int right)
{
if (left>right) return 0;
if (left<=tree[cur].left && tree[cur].right<=right)
{
return tree[cur].sum;
}
int mid=(tree[cur].left+tree[cur].right)>>1;
int res1=0,res2=0;
if (left<=mid) res1=query(cur*2,left,right);
if (mid+1<=right) res2=query(cur*2+1,left,right);
return res1+res2;
}
int n;
Pair a[100048];
int pos[100048];
bool cmp(int x,int y)
{
if (x-stand>0 && y-stand<0 || x-stand<0 && y-stand>0) return x-stand>y-stand;
if (x-stand>0 && y-stand>0) return x-stand<y-stand;
if (x-stand<0 && y-stand<0) return x-stand<y-stand;
}
int main ()
{
int i,j;
scanf("%d",&n);
for (i=1;i<=n;i++)
{
scanf("%d",&a[i].x);
a[i].y=i;
v[a[i].x].pb(i);
}
int t=1;
while (v[t].size()==0) t++;
sort(v[t].begin(),v[t].end());
stand=v[t].back();
t++;
do
{
while (t<=100000 && v[t].size()==0) t++;
if (t>100000) break;
sort(v[t].begin(),v[t].end(),cmp);
stand=v[t].back();
t++;
}
while (t<=100000);
build(1,1,n);
int top=0;
for (i=1;i<=100000;i++)
if (v[i].size()!=0)
for (j=0;j<v[i].size();j++)
pos[++top]=v[i][j];
pos[0]=0;
LL ans=0;
int cnt;
for (i=1;i<=n;i++)
{
if (pos[i-1]+1<=pos[i])
{
cnt=query(1,pos[i-1]+1,pos[i]);
ans+=cnt;
}
else
{
cnt=query(1,pos[i-1]+1,n);
cnt+=query(1,1,pos[i]);
ans+=cnt;
}
update(1,pos[i]);
}
cout<<ans<<endl;
return 0;
}