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  • UOJ

    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
    ", query(1, 1, n, l, r));
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/Pneuis/p/9026039.html
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