LOJ 数列分块入门6
题目:
题解:
- 我都不懂这题为什么要用分块... ...
- 直接vector就好了...
- 但是如果有区间修改的话就不行了。所以这题是启示我们也可以动态分块。具体就是每次插入时先找到位置所在的块,再暴力插入,把块内的其它元素直接向后移动一位。如果对于一个块插入过多元素时,需要重构(重新分块),从而使分块拥有更高的性能。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#define N 2000005
using namespace std;
vector<int> a;
struct Blo {int l, r;} blo[N];
int n, num, size;
int bel[N];
int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return x *= f;
}
void build(int len) {
size = (int)sqrt(len), num = len / size;
if (len % size) num++;
for (int i = 1; i <= len; i++) bel[i] = (i - 1) / size + 1;
for (int i = 1; i <= num; i++) {
blo[i].l = (i - 1) * size;
blo[i].r = i * size;
}
blo[num].r = len;
}
void update(int pos, int val) {
blo[bel[pos]].r++;
for (int i = bel[pos] + 1; i <= num; i++) {
blo[i].l++;
blo[i].r++;
}
a.insert(a.begin() + pos, val);
}
int ask(int pos) {
return a[pos];
}
int main() {
cin >> n;
a.push_back(-1);
for (int i = 1; i <= n; i++) a.push_back(read());
build(n);
for (int i = 1; i <= n; i++) {
int op = read(), l = read(), r = read(), c = read();
if(!op) update(l, r);
else printf("%d\n", ask(r));
}
return 0;
}