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+;
 const int MAXV = ;
 const int MAXE = ;
 const ll MOD = ;
 ll sum1[MAXN << ];
 ll sum2[MAXN << ];
 ll sum3[MAXN << ];
 ll addmark[MAXN << ];
 ll setmark[MAXN << ];
 ll mulmark[MAXN << ];

 void pushDown(int root, int l, int r)
 {
     int num = (r-l+);
     ll numr = num >> ;
     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)] = ;
         mulmark[lson(root)] = mulmark[rson(root)] = ;
         setmark[root] = ;
     }
     if (mulmark[root] != )
     {
         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] = ;
     }
     if (addmark[root])
     {
         ll addval = addmark[root];
         sum3[lson(root)] = ((sum3[lson(root)]+numl*addval*addval*addval)%MOD+(*sum2[lson(root)]*addval+*sum1[lson(root)]*addval*addval)%MOD)%MOD;
         sum3[rson(root)] = ((sum3[rson(root)]+numr*addval*addval*addval)%MOD+(*sum2[rson(root)]*addval+*sum1[rson(root)]*addval*addval)%MOD)%MOD;
         sum2[lson(root)] = (((sum2[lson(root)] + numl*addval*addval) % MOD) + (*sum1[lson(root)]*addval)) % MOD;
         sum2[rson(root)] = (((sum2[rson(root)] + numr*addval*addval) % MOD) + (*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] = ;
     }
 }
 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+;
         if (mathod == )
         {
             //cout << l << " " << r << " " << val << endl;
             setmark[root] = val%MOD;
             addmark[root] = ;
             mulmark[root] = ;
             sum1[root] = (num*val) % MOD;
             sum2[root] = (num*val*val) % MOD;
             sum3[root] = (num*val*val*val) % MOD;
         }
         else if (mathod == )
         {
             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 == )
         {

             //cout << l << " "  << r << " " << sum1[root] << endl;
             addmark[root] = (addmark[root]+val)%MOD;
             sum3[root] = ((sum3[root]+num*val*val*val)%MOD+(*sum2[root]*val+*sum1[root]*val*val)%MOD)%MOD;
             sum2[root] = ((sum2[root] + num*val*val) % MOD + (*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) >> ;
     if (ul <= mid) update(lson(root), l, mid, ul, ur, mathod, val);
     if (ur > mid) update(rson(root), mid+, 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 = ;
 ll query(int root, int l, int r, int ql, int qr, int p)
 {
    // cout << l << " " << r << endl;
     if (l > qr || r < ql) return ;
     if (l >= ql && r <= qr)
     {
         if (p == ) return sum1[root];
         else if (p == ) return sum2[root];
         else if (p == ) return sum3[root];
     }
     pushDown(root, l, r);
     int mid = (l+r) >> ;
     ll res = ;
     if (ql <= mid) res = (res + query(lson(root), l, mid, ql, qr, p)) % MOD;
     if (qr >= mid+) res = (res + query(rson(root), mid+, 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 = ; i < MAXN << ; i++) mulmark[i] = ;
         if (!n && !m) break;
         memset(addmark, , sizeof(addmark));
         //memset(mulmark, 1, sizeof(mulmark));
         memset(setmark, , sizeof(setmark));
         memset(sum1, , sizeof(sum1));
         memset(sum2, , sizeof(sum2));
         memset(sum3, , sizeof(sum3));
         for (int i = ; i < m; i++)
         {
             int op, l, r, c;
             scanf("%d%d%d%d", &op, &l, &r, &c);
             if (op == )
             {
                 update(, , n, l, r, , c);
             }
             else if (op == )
             {
                 update(, , n, l, r, , c);
             }
             else if (op == )
             {
                 update(, , n, l, r, , c);
             }
             else
             {
                 ll ans = query(, , n, l, r, c);
                 cout << ans << endl;
             }
         }
     }
     return ;
 }