链接:
题意:
给出一个长为 的数列,以及 个操作,操作涉及单点插入,单点询问,数据随机生成。
思路:
vector 维护每个区间, 当某个区间的值太多时,重构一下.
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;
int Belong[MAXN*10], a[MAXN*10];
vector<int> Vec[MAXN];
int n, part, last;
pair<int, int> GetPos(int p)
{
int p1 = 1;
while (p > Vec[p1].size())
p -= Vec[p1].size(), p1++;
return make_pair(p1, p-1);
}
void Rebuild()
{
int pos = 0;
for (int i = 1;i <= last;i++)
{
for (vector<int>::iterator it = Vec[i].begin();it != Vec[i].end();++it)
a[++pos] = *it;
Vec[i].clear();
}
part = sqrt(pos);
last = (pos-1)/part+1;
for (int i = 1;i <= pos;i++)
{
Belong[i] = (i-1)/part+1;
Vec[Belong[i]].push_back(a[i]);
}
}
void Update(int l, int r, int c)
{
pair<int, int> p = GetPos(l);
Vec[p.first].insert(Vec[p.first].begin()+p.second, r);
if (Vec[p.first].size() > 10*part)
Rebuild();
}
int Query(int l, int r, int c)
{
pair<int, int> p = GetPos(r);
return Vec[p.first][p.second];
}
int main()
{
scanf("%d", &n);
part = sqrt(n);
int v;
for (int i = 1;i <= n;i++)
{
scanf("%d", &v);
Belong[i] = (i-1)/part+1;
Vec[Belong[i]].push_back(v);
}
last = (n-1)/part+1;//最后一个区间
int op, l, r, c;
for (int i = 1;i <= n;i++)
{
scanf("%d", &op);
if (op == 0)
{
scanf("%d%d%d", &l, &r, &c);
Update(l, r, c);
}
else
{
scanf("%d%d%d", &l, &r, &c);
printf("%d
", Query(l, r, c));
}
}
return 0;
}