线段树懒标记好题 HDU4578

（1）"1 x y c",代表 把区间 [x,y] 上的值全部加c

（2）"2 x y c",代表 把区间 [x,y] 上的值全部乘以c

（3）"3 x y c" 代表 把区间 [x,y]上的值全部赋值为c

（4）"4 x y p" 代表 求区间 [x,y] 上值的p次方和1<=p<=3

维护sum1[], sum2[], sum3[]分别为一次方、二次方、三次方的和

      因为P只有1到3，(x+c)^2=(x^2)+(2*x*c+c^2)，即我们可以从一次方和的结果推出二次方的和的结果。同理，我们也可以根据一次方的和的结果，二次方和的结果推出三次方和的结果。这样，我们可以用几个懒标记去记录这几个操作。但是这题明白上面这点还是不够的，因为，这几种更新操作之间会相互影响，比如对一个区间进行操作3之后，那么操作1和2就失效了。因此在PushDown传递懒标记的时候，传递的顺序也是重要的。

#include <bits/stdc++.h>
#include <string.h>
#include <iostream>
#include <stdio.h>
#define pb push_back
#define fi first
#define se second
#define lson(r) (r<<1)
#define rson(r) ((r<<1)|1)

using namespace std;

typedef long long ll;

const int MAXN = 1e5+7;
const int MAXV = 507;
const int MAXE = 507;
const ll MOD = 10007;
ll sum1[MAXN << 2];
ll sum2[MAXN << 2];
ll sum3[MAXN << 2];
ll addmark[MAXN << 2];
ll setmark[MAXN << 2];
ll mulmark[MAXN << 2];

void pushDown(int root, int l, int r)
{
    int num = (r-l+1);
    ll numr = num >> 1;
    ll numl = num - numr;
    //思考了半天用-1还是不够优秀呀 只需要把另外优先级低的懒标记 标记的左右子即可 
    //修改了半天 哎哎哎 这道题目深入理解懒标记
    if (setmark[root])
    {
        ll setval = setmark[root];
        //cout << "set" << l << " " << r << " " << setmark[root] << endl;
        sum1[lson(root)] = (numl*setval) % MOD;
        sum1[rson(root)] = (numr*setval) % MOD;
        sum2[lson(root)] = (numl*setval*setval) % MOD;
        sum2[rson(root)] = (numr*setval*setval) % MOD;
        sum3[lson(root)] = (numl*setval*setval*setval) % MOD;
        sum3[rson(root)] = (numr*setval*setval*setval) % MOD;
        setmark[lson(root)] = setval;
        setmark[rson(root)] = setval;
        addmark[lson(root)] = addmark[rson(root)] = 0;
        mulmark[lson(root)] = mulmark[rson(root)] = 1;
        setmark[root] = 0;
    }
    if (mulmark[root] != 1)
    {
        ll mulval = mulmark[root];
        sum1[lson(root)] = (sum1[lson(root)]*mulval) % MOD;
        sum1[rson(root)] = (sum1[rson(root)]*mulval) % MOD;
        sum2[lson(root)] = (sum2[lson(root)]*mulval*mulval) % MOD;
        sum2[rson(root)] = (sum2[rson(root)]*mulval*mulval) % MOD;
        sum3[lson(root)] = (sum3[lson(root)]*mulval*mulval*mulval) % MOD;
        sum3[rson(root)] = (sum3[rson(root)]*mulval*mulval*mulval) % MOD;
        mulmark[lson(root)] = (mulmark[lson(root)]*mulval) % MOD;
        mulmark[rson(root)] = (mulmark[rson(root)]*mulval) % MOD;
        addmark[lson(root)] = (addmark[lson(root)]*mulval) % MOD;
        addmark[rson(root)] = (addmark[rson(root)]*mulval) % MOD;
        mulmark[root] = 1;
    }    
    if (addmark[root])
    {
        ll addval = addmark[root];
        sum3[lson(root)] = ((sum3[lson(root)]+numl*addval*addval*addval)%MOD+(3*sum2[lson(root)]*addval+3*sum1[lson(root)]*addval*addval)%MOD)%MOD;
        sum3[rson(root)] = ((sum3[rson(root)]+numr*addval*addval*addval)%MOD+(3*sum2[rson(root)]*addval+3*sum1[rson(root)]*addval*addval)%MOD)%MOD;
        sum2[lson(root)] = (((sum2[lson(root)] + numl*addval*addval) % MOD) + (2*sum1[lson(root)]*addval)) % MOD;
        sum2[rson(root)] = (((sum2[rson(root)] + numr*addval*addval) % MOD) + (2*sum1[rson(root)]*addval)) % MOD;
        sum1[lson(root)] = (sum1[lson(root)] + numl * addval) % MOD;
        sum1[rson(root)] = (sum1[rson(root)] + numr * addval) % MOD;
        addmark[lson(root)] = (addmark[lson(root)] + addval) % MOD;
        addmark[rson(root)] = (addmark[rson(root)] + addval) % MOD;
        addmark[root] = 0;
    }
}
void update(int root, int l, int r, int ul, int ur, int mathod, ll val)
{
    if (l > ur || r < ul) return ;
    if (l >= ul && r <= ur)
    {
        ll num = r-l+1;
        if (mathod == 1)
        {
            //cout << l << " " << r << " " << val << endl;
            setmark[root] = val%MOD;
            addmark[root] = 0;
            mulmark[root] = 1;
            sum1[root] = (num*val) % MOD;
            sum2[root] = (num*val*val) % MOD;
            sum3[root] = (num*val*val*val) % MOD;
        }
        else if (mathod == 2)
        {
            mulmark[root] = (mulmark[root]*val)%MOD;
            addmark[root] = (val*addmark[root])%MOD;
            sum1[root] = (sum1[root]*val) % MOD;
            sum2[root] = (sum2[root]*val*val) % MOD;
            sum3[root] = (sum3[root]*val*val*val) % MOD;
        }
        else if (mathod == 3)
        {

            //cout << l << " "  << r << " " << sum1[root] << endl;
            addmark[root] = (addmark[root]+val)%MOD;
            sum3[root] = ((sum3[root]+num*val*val*val)%MOD+(3*sum2[root]*val+3*sum1[root]*val*val)%MOD)%MOD;
            sum2[root] = ((sum2[root] + num*val*val) % MOD + (2*sum1[root]*val)) % MOD;
            sum1[root] = (sum1[root] + num * val) % MOD;
            //cout << l << " "  << r << " " << sum1[root] << endl;
        }
        return ;
    }
    pushDown(root, l, r);
    int mid = (l+r) >> 1;
    if (ul <= mid) update(lson(root), l, mid, ul, ur, mathod, val);
    if (ur > mid) update(rson(root), mid+1, r, ul, ur, mathod, val);
    sum1[root] = (sum1[lson(root)] + sum1[rson(root)]) % MOD;
    sum2[root] = (sum2[lson(root)] + sum2[rson(root)]) % MOD;
    sum3[root] = (sum3[lson(root)] + sum3[rson(root)]) % MOD;
    //cout << l << " " << r << " " << sum1[root] << endl;
}

int cnt = 0;
ll query(int root, int l, int r, int ql, int qr, int p)
{
   // cout << l << " " << r << endl;
    if (l > qr || r < ql) return 0;
    if (l >= ql && r <= qr)
    {
        if (p == 1) return sum1[root];
        else if (p == 2) return sum2[root];
        else if (p == 3) return sum3[root];
    }
    pushDown(root, l, r);
    int mid = (l+r) >> 1;
    ll res = 0;
    if (ql <= mid) res = (res + query(lson(root), l, mid, ql, qr, p)) % MOD;
    if (qr >= mid+1) res = (res + query(rson(root), mid+1, r, ql, qr, p)) % MOD; 
    return res;
}
int n, m;
int main()
{
   // freopen("in.txt", "r", stdin);
    while (~scanf("%d%d", &n, &m))
    {
        for(int i = 0; i < MAXN << 2; i++) mulmark[i] = 1;
        if (!n && !m) break;
        memset(addmark, 0, sizeof(addmark));
        //memset(mulmark, 1, sizeof(mulmark));
        memset(setmark, 0, sizeof(setmark));
        memset(sum1, 0, sizeof(sum1));
        memset(sum2, 0, sizeof(sum2));
        memset(sum3, 0, sizeof(sum3));
        for (int i = 0; i < m; i++)
        {
            int op, l, r, c;
            scanf("%d%d%d%d", &op, &l, &r, &c);
            if (op == 1) 
            {
                update(1, 1, n, l, r, 3, c);            
            }
            else if (op == 2) 
            {
                update(1, 1, n, l, r, 2, c);
            }
            else if (op == 3) 
            {
                update(1, 1, n, l, r, 1, c);
            }
            else 
            {
                ll ans = query(1, 1, n, l, r, c);
                cout << ans << endl;
            }
        }
    }
    return 0;
}