LOJ 6278 数列分块入门 2

嗯...

题目链接：https://loj.ac/problem/6278

这道题只要加上一个update函数即可，用一个vector来记录当前的元素，分三种情况：

1.在l的块中：直接暴力枚举

2.在r的块中：直接暴力枚举

3.在整块中：二分答案

AC代码：

#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>

using namespace std;

const int maxn = 100010;
const int N = 505;

int block, num;
int L[N], R[N], add[N];
int a[maxn], belong[maxn];
vector<int> vc[N];

inline void build(int n){
    block = sqrt(n);
    num = ceil(n * 1.0 / block);
    for(int i = 1; i <= n; i++){
        belong[i] = (i - 1) / block + 1;
        vc[belong[i]].push_back(a[i]);
    }
    for(int i = 1; i <= num; i++){
        L[i] = (i - 1) * block + 1;
        R[i] = i * block;
        sort(vc[i].begin(), vc[i].end());
    }
    R[num] = n;
}

inline void update(int pos){
    vc[pos].clear();
    for(int i = L[pos]; i <= R[pos]; i++) vc[pos].push_back(a[i]);
    sort(vc[pos].begin(), vc[pos].end());
}

inline void modify(int l, int r, int c){
    for(int i = l; i <= min(r, R[belong[l]]); i++) a[i] += c;
    update(belong[l]);
    if(belong[l] != belong[r]){
        for(int i = L[belong[r]]; i <= r; i++) a[i] += c;
        update(belong[r]);
    }
    for(int i = belong[l] + 1; i < belong[r]; i++) add[i] += c;
}

inline int query(int l, int r, int c){
    int ans = 0;
    for(int i = l; i <= min(r, R[belong[l]]); i++)
        if(a[i] + add[belong[i]] < c) ans++;
    if(belong[l] != belong[r]){
        for(int i = L[belong[r]]; i <= r; i++)
            if(a[i] + add[belong[i]] < c) ans++;
    }
    for(int i = belong[l] + 1; i < belong[r]; i++)
        ans += lower_bound(vc[i].begin(), vc[i].end(), c - add[i]) - vc[i].begin();
    return ans;
}

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(n);
    for(int i = 1; i <= n; i++){
        int opt, l, r, c;
        scanf("%d%d%d%d", &opt, &l, &r, &c);
        if(opt == 0) modify(l, r, c);
        else printf("%d
", query(l, r, c * c));
    }
    return 0;
}