[洛谷P2801]教主的魔法

题目传送门

一道分块的好题。

这题分块后，对于两种操作：

·让区间[l,r]+=w；

·查询区间[l,r]>=c的数的个数

分块后，我们将每一块中的数排序，这样每一块中的查询可以通过二分完成。

对于区间的修改，如果覆盖了整块，通过标签的修改满足题意；如果只是块中的一部分，暴力修改原数组再重新排序维护。

第一个复杂度为$O(sqrt{n})$，第二个为$O(sqrt{n}log sqrt{n})$。总的为$O(Qsqrt{n}log sqrt{n})$

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define inf (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 1e6 + 5, Size = 1000;

int A[maxn], B[maxn], N, Q, tag[maxn];

#define pos(x) ((x+Size-1)/Size)

void Build(int P) {
    rep(i, (P-1)*Size+1, P*Size) B[i] = A[i];
    sort(B + (P-1)*Size + 1, B + P*Size + 1);
}

void Update(int L, int R, int W) {
    int Left = pos(L)+1, Right = pos(R)-1;
    rep(i, Left, Right) tag[i] += W;
    if (Left-1 > Right) {
        rep(i, L, R) A[i] += W;
        Build(Left-1);
    } else {
        rep(i, L, (Left-1)*Size) A[i] += W;
        rep(i, Right*Size+1, R) A[i] += W;
        Build(Left-1), Build(Right+1);
    }
}

int find(int P, int v) {
    register int L = (P-1)*Size+1, R = P*Size, Mid, Ans = R+1;
    while (L <= R) {
        Mid = (L + R) >> 1;
        if (B[Mid] >= v) R = Mid-1, Ans = Mid;
        else L = Mid+1;
    }
    return P*Size-Ans+1;
}

int Query(int L, int R, int C) {
    int Left = pos(L)+1, Right = pos(R)-1, Tot = 0;
    rep(i, Left, Right) Tot += find(i, C-tag[i]);
    if (Left-1 > Right) {
        rep(i, L, R) if (A[i] >= C-tag[Left-1]) Tot++;
    } else {
        rep(i, L, (Left-1)*Size) if (A[i] >= C-tag[Left-1]) Tot++;
        rep(i, Right*Size+1, R) if (A[i] >= C-tag[Right+1]) Tot++;
    }
    return Tot;
}

int main() {
    N = read(), Q = read();
    rep(i, 1, N) B[i] = A[i] = read();

    for (register int i = 1; i <= N; i += Size) sort(B + i, B + min(N+1, Size + i));
    memset(tag, 0, sizeof(tag));

    rep(i, 1, Q) {
        char c = getchar();
        while (c != 'A' && c != 'M') c = getchar();
        int L = read(), R = read(), W = read();
        if (c == 'A') printf("%d
", Query(L, R, W));
        else if (c == 'M') Update(L, R, W);
    }
    return 0;
}

当然还有更优的做法。这里就不放了。