UOJ - #228. 基础数据结构练习题

题意：

　　一个区间支持三种操作，区间加，区间开根号和区间求和。

题解：

　　线段树的做法。对于区间开根号操作，如果要开根号的区间最大值和最小值相等的话相当于区间减操作。当最大值和最小值相差1时，如果最大值是平方数那么也相当于区间减操作，否则就是区间覆盖。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define lson (id<<1)
#define rson ((id<<1)|1)
#define mid ((l+r)>>1)
using namespace std;
typedef long long ll;
const int N = 1e5+10;
int n, m;
int t, l, r;
ll x;
ll lazy[N*5], maxx[N*5], minn[N*5], sum[N*5];
void push_up(int id) {
    maxx[id] = max(maxx[lson], maxx[rson]);
    minn[id] = min(minn[lson], minn[rson]);
    sum[id] = sum[lson]+sum[rson];
}
void build(int id, int l, int r) {
    lazy[id] = 0;
    if(l==r) {
        scanf("%lld", &sum[id]);
        maxx[id] = minn[id] = sum[id];
        return ;
    }
    build(lson, l, mid);
    build(rson, mid+1, r);
    push_up(id);
    return ;
}
void push_down(int id, int l, int r) {
    if(lazy[id]) {
        lazy[lson] += lazy[id];
        lazy[rson] += lazy[id];
        sum[lson] += 1ll*(mid-l+1)*lazy[id];
        sum[rson] += 1ll*(r-mid)*lazy[id];
        maxx[lson] += lazy[id];
        maxx[rson] += lazy[id];
        minn[lson] += lazy[id];
        minn[rson] += lazy[id];
        lazy[id] = 0;
    }
}
void ins(int id, int l, int r, int ql, int qr, ll x) {
    if(ql<=l&&r<=qr) {
        sum[id] += 1ll*(r-l+1)*x;
        lazy[id] += x;
        maxx[id] += x;
        minn[id] += x;
        return ;
    }
    push_down(id, l, r);
    if(ql<=mid) ins(lson, l, mid, ql, qr, x);
    if(qr>mid) ins(rson, mid+1, r, ql, qr, x);
    push_up(id);
}
void update(int id, int l, int r, int ql, int qr) {
    if(ql<=l&&r<=qr&&maxx[id]-minn[id]<=1) {
        ll k = (ll)sqrt(maxx[id]);
        if(maxx[id]-minn[id]==0||k*k==maxx[id]) {
            ins(id, l, r, l, r, k-maxx[id]);
            return ;
        }
    }
    push_down(id, l, r);
    if(ql<=mid) update(lson, l, mid, ql, qr);
    if(qr>mid) update(rson, mid+1, r, ql, qr);
    push_up(id);
}
ll query(int id, int l, int r, int ql, int qr) {
    if(ql<=l&&r<=qr) return sum[id];
    push_down(id, l, r);
    ll res = 0;
    if(ql<=mid) res += query(lson, l, mid, ql, qr);
    if(qr>mid) res += query(rson, mid+1, r, ql, qr);
    push_up(id);
    return res;
}
int main() {
    scanf("%d%d", &n, &m);
    build(1, 1, n);
    while(m--) {
        scanf("%d%d%d", &t, &l, &r);
        if(t==1) {
            scanf("%lld", &x);
            ins(1, 1, n, l, r, x);
        }
        if(t==2) update(1, 1, n, l, r);
        if(t==3) printf("%lld\n", query(1, 1, n, l, r));
    }
}

原文地址：https://www.cnblogs.com/Pneuis/p/9026039.html