数列分块入门 2

分块入门

每个块多用个vector排序即可

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 50005;
ll a[N], add[N];
int n, lt[N], rt[N], pos[N], t;
vector<ll> v[N];

void modify(int l, int r, ll d){
    int p = pos[l], q = pos[r];
    if(p == q){
        v[p].clear();
        for(int i = l; i <= r; i ++) a[i] += d;
        for(int i = lt[p]; i <= rt[p]; i ++) v[p].push_back(a[i]);
        sort(v[p].begin(), v[p].end());
    }
    else{
        for(int i = p + 1; i <= q - 1; i ++) add[i] += d;
        v[p].clear();
        for(int i = l; i <= rt[p]; i ++) a[i] += d;
        for(int i = lt[p]; i <= rt[p]; i ++) v[p].push_back(a[i]);
        sort(v[p].begin(), v[p].end());
        v[q].clear();
        for(int i = lt[q]; i <= r; i ++) a[i] += d;
        for(int i = lt[q]; i <= rt[q]; i ++) v[q].push_back(a[i]);
        sort(v[q].begin(), v[q].end());
    }
}

int query(int l, int r, ll k){
    int p = pos[l], q = pos[r], ret = 0;
    ll tmp;
    if(p == q){
        tmp = k - add[p];
        for(int i = l; i <= r; i ++){
            if(a[i] < tmp) ret ++;
        }
    }
    else{
        for(int i = p + 1; i <= q - 1; i ++){
            tmp = k - add[i];
            ret += lower_bound(v[i].begin(), v[i].end(), tmp) - v[i].begin();
        }
        tmp = k - add[p];
        for(int i = l; i <= rt[p]; i ++){
            if(a[i] < tmp) ret ++;
        }
        tmp = k - add[q];
        for(int i = lt[q]; i <= r; i ++){
            if(a[i] < tmp) ret ++;
        }
    }
    return ret;
}

int main(){
    //freopen("data.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    t = (int)sqrt(n);
    for(int i = 1; i <= t; i ++){
        lt[i] = (i - 1) * t + 1;
        rt[i] = i * t;
    }
    if(rt[t] < n) t ++, lt[t] = rt[t - 1] + 1, rt[t] = n;
    for(int i = 1; i <= t; i ++){
        for(int j = lt[i]; j <= rt[i]; j ++){
            pos[j] = i;
            v[i].push_back(a[j]);
        }
        sort(v[i].begin(), v[i].end());
    }
    for(int i = 1; i <= n; i ++){
        int opt, l, r; ll c;
        cin >> opt >> l >> r >> c;
        if(!opt) modify(l, r, c);
        else cout << query(l, r, c * c) << endl;
    }
    return 0;
}

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转载自www.cnblogs.com/onionQAQ/p/10872531.html
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