參考：「分塊」數列分塊入門1 – 9 by hzwer

2

Description

給出一個長為\(n\)的數列，以及\(n\)個操作，操作涉及區間加法，詢問區間內小於某個值\(x\)的元素個數。

思路

每個塊內保持升序排列。

則塊外暴力統計，塊內二分查找分界點。

一些註意點，如：

1. 要記錄下標；
2. 塊外暴力修改完之後需要再排序；
3. 在塊內二分查找的值是\(c-tag[i]\)而非\(c\).

Code

#include <bits/stdc++.h>
#define maxn 50010
#define F(i, a, b) for (int i = (a); i < (b); ++i)
 

#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long LL;
int tag[maxn], bl[maxn], n, blo;
struct node {
    int x, p;
    bool operator < (const
 
 node& nd) const { return x < nd.x; }
}a[maxn];
inline int val(int x) { return a[x].x + tag[bl[x]]; }
int query(int l, int r, int c) {
    int ret=0;
    F(i, bl[l]*blo, min((bl[l]+1)*blo, n)) if (a[i].p>=l&&a[i].p<=r && val(i)<c) ++ret;
    if (bl[l]!=bl[r]) F(i, bl[r]*blo, min((bl[r]+1
 
)*blo, n)) if (a[i].p>=l&&a[i].p<=r && val(i)<c) ++ret;
    F(i, bl[l]+1, bl[r]) ret += lower_bound(a+i*blo, a+(i+1)*blo, (node){c-tag[i], 0}) - (a+i*blo);
    return ret;
}
void add(int l, int r, int c) {
    F(i, bl[l]*blo, min((bl[l]+1)*blo,n)) if (a[i].p>=l&&a[i].p<=r) a[i].x+=c;
    sort(a+bl[l]*blo, a+min((bl[l]+1)*blo, n));
    if (bl[l]!=bl[r]) {
        F(i, bl[r]*blo, min((bl[r]+1)*blo, n)) if (a[i].p>=l&&a[i].p<=r) a[i].x+=c;
        sort(a+bl[r]*blo, a+min((bl[r]+1)*blo, n));
    }
    F(i, bl[l]+1, bl[r]) tag[i] += c;
}
int main() {
    scanf("%d", &n); blo = sqrt(n);
    F(i, 0, n) scanf("%d", &a[i].x), a[i].p = i, bl[i] = i/blo;
    int num = (n+blo-1)/blo;
    F(i, 0, num-1) sort(a+i*blo, a+(i+1)*blo);
    sort(a+(num-1)*blo, a+n);
    F(i, 0, n) {
        int op, l, r, c;
        scanf("%d%d%d%d", &op, &l, &r, &c); --l, --r;
        if (op) printf("%d\n", query(l, r, c*c));
        else add(l, r, c);
    }
    return 0;
}

3

Description

給出一個長為\(n\)的數列，以及\(n\)個操作，操作涉及區間加法，詢問區間內小於某個值\(x\)的最大值。

思路

做法基本同上。

Code

#include <bits/stdc++.h>
#define maxn 100010
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long LL;
int tag[maxn], bl[maxn], n, blo;
struct node {
    int x, p;
    bool operator < (const node& nd) const { return x < nd.x; }
}a[maxn];
inline int val(int x) { return a[x].x + tag[bl[x]]; }
int query(int l, int r, int c) {
    int ans=-1, diff=INT_MAX, temp;
    F(i, bl[l]*blo, min((bl[l]+1)*blo, n)) {
        if (a[i].p>=l&&a[i].p<=r && (temp=c-val(i))>0 && temp<diff) ans = val(i), diff = temp;
    }
    if (bl[l]!=bl[r]) F(i, bl[r]*blo, min((bl[r]+1)*blo, n)) {
        if (a[i].p>=l&&a[i].p<=r && (temp=c-val(i))>0 && temp<diff) ans = val(i), diff = temp;
    }
    F(i, bl[l]+1, bl[r]) {
        int p = lower_bound(a+i*blo, a+(i+1)*blo, (node){c-tag[i], 0}) - (a+i*blo);
        if (p==0) continue;
        temp = val(i*blo+p-1);
        if (c-temp>0 && c-temp<diff) diff = c-temp, ans = temp;
    }
    return ans;
}
void add(int l, int r, int c) {
    F(i, bl[l]*blo, min((bl[l]+1)*blo,n)) if (a[i].p>=l&&a[i].p<=r) a[i].x+=c;
    sort(a+bl[l]*blo, a+min((bl[l]+1)*blo, n));
    if (bl[l]!=bl[r]) {
        F(i, bl[r]*blo, min((bl[r]+1)*blo, n)) if (a[i].p>=l&&a[i].p<=r) a[i].x+=c;
        sort(a+bl[r]*blo, a+min((bl[r]+1)*blo, n));
    }
    F(i, bl[l]+1, bl[r]) tag[i] += c;
}
int main() {
    scanf("%d", &n); blo = sqrt(n);
    F(i, 0, n) scanf("%d", &a[i].x), a[i].p = i, bl[i] = i/blo;
    int num = (n+blo-1)/blo;
    F(i, 0, num-1) sort(a+i*blo, a+(i+1)*blo);
    sort(a+(num-1)*blo, a+n);
    F(i, 0, n) {
        int op, l, r, c;
        scanf("%d%d%d%d", &op, &l, &r, &c); --l, --r;
        if (op) printf("%d\n", query(l, r, c));
        else add(l, r, c);
    }
    return 0;
}

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