LOJ6279 数列分块入门3
LOJ6279 数列分块入门 3
标签
- 分块入门
前言
- 犯了一些低级错误,debug了半天~
简明题意
- 维护序列,支持两种操作:
- 区间加
- 查询某个数的前驱
思路
- 这题和数列分块入门 2基本差不多了,就是查询的时候稍微改变一下就可以了QAQ
- 具体就是整块的直接二分,不整块的暴力找,然后取最大的就行了
注意事项
- 一定要注意细节。tag不要忘记加了
总结
- 分块好玩又简单QAQ,为啥没有早点学呢
AC代码
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int maxn = 1e5 + 10;
int n, a[maxn];
int pos[maxn], len, tag[maxn];
vector<int> b[maxn];
void reset(int id)
{
b[id].clear();
for (int i = (id - 1) * len + 1; i <= min(id * len, n); i++)
b[id].push_back(a[i]);
sort(b[id].begin(), b[id].end());
}
void change(int l, int r, int c)
{
for (int i = l; i <= min(pos[l] * len, r); i++)
a[i] += c;
reset(pos[l]);
if (pos[l] != pos[r])
{
for (int i = r; i >= (pos[r] - 1) * len + 1; i--)
a[i] += c;
reset(pos[r]);
}
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
tag[i] += c;
}
int cal(int l, int r, int c)
{
int ans = -1;
for (int i = l; i <= min(pos[l] * len, r); i++)
if (a[i] + tag[pos[i]] < c)
ans = max(ans, a[i] + tag[pos[i]]);
if (pos[l] != pos[r])
for (int i = r; i >= max((pos[r] - 1) * len + 1, l); i--)
if (a[i] + tag[pos[i]] < c)
ans = max(ans, a[i] + tag[pos[i]]);
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
{
auto t = lower_bound(b[i].begin(), b[i].end(), c - tag[i]);
if (t != b[i].begin() && *(t - 1) + tag[i] < c)
ans = max(ans, *(t - 1) + tag[i]);
}
return ans;
}
void solve()
{
scanf("%d", &n);
len = sqrt(n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]), b[pos[i] = (i - 1) / len + 1].push_back(a[i]);
//预处理
for (int i = 1; i <= pos[n]; i++)
sort(b[i].begin(), b[i].end());
for (int i = 1; i <= n; i++)
{
int opt, l, r, c;
scanf("%d%d%d%d", &opt, &l, &r, &c);
if (opt == 0)
change(l, r, c);
else
printf("%d\n", cal(l, r, c));
}
}
int main()
{
freopen("Testin.txt", "r", stdin);
freopen("Testout.txt", "w", stdout);
solve();
return 0;
}
作者:danzh
QQ:1244536605
CSDN(和博客园同步):https://blog.csdn.net/weixin_42431507
-----------------------------------------------------------------------------------------------
朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其
它人怎么样,我们也能够保持自己的本色走下去。
—clj