HDU4578 线段树(区间更新 + 多种操作)和平方，立方
参考：https://www.cnblogs.com/H-Vking/p/4297973.html
题意：
虽然是比较裸的线段树，但是比较麻烦，并且有很多细节需要考虑，对着别人的ac代码debug了一个晚上。纪念一下
这道题坑在有三种询问：set , add , mul。所以lazy标记要有三个，如果三个标记同时出现的处理方法——当更新set操作时，就把add标记和mul标记全部取消；当更新mul操作时，如果当前节点add标记存在，就把add标记改为：add * mul。这样的话就可以在PushDown()操作中先执行set，然后mul，最后add。
麻烦在有三种询问：和 , 平方和 , 立方和。对于set和mul操作来说，这三种询问都比较好弄。
对于add操作，和的话就比较好弄，按照正常方法就可以；
平方和这样来推：(a + c)2 = a2 + c2 + 2ac , 即sum2[rt] = sum2[rt] + (r - l + 1) * c * c + 2 * sum1[rt] * c；
立方和这样推：(a + c)3 = a3 + c3 + 3a(a2 + ac) , 即sum3[rt] = sum3[rt] + (r - l + 1) * c * c * c + 3 * c * (sum2[rt] + sum1[rt] * c)；
几个注意点：add标记取消的时候是置0，mul标记取消的时候是置1；在PushDown()中也也要注意取消标记，如set操作中取消add和mul，mul操作中更新add； 在add操作中要注意sum3 , sum2 , sum1的先后顺序，一定是先sum3 , 然后sum2 , 最后sum1； int容易爆，还是用LL要保险一点； 最后就是运算较多，不要漏掉东西。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <list> #include <iterator> using namespace std; #define lson l , mid , rt << 1 #define rson mid + 1 , r , rt << 1 | 1 #define debug(x) cerr << #x << " = " << x << "\n"; typedef long long ll; const int maxn = 1e5+9; const int mdd = 10007; struct node { int l,r; ll sum1,sum2,sum3; ll lazy1,lazy2,lazy3; }st[maxn<<2]; void allmdd(int rt) { st[rt].sum1 = st[rt].sum1%mdd; st[rt].sum2 = st[rt].sum2%mdd; st[rt].sum3 = st[rt].sum3%mdd; } void pushup(int rt) { st[rt].sum1 = (st[rt<<1].sum1 + st[rt<<1|1].sum1)%mdd; st[rt].sum2 = (st[rt<<1].sum2 + st[rt<<1|1].sum2)%mdd; st[rt].sum3 = (st[rt<<1].sum3 + st[rt<<1|1].sum3)%mdd; } void pushdown(int rt) { int len = st[rt].r - st[rt].l + 1; int llen = len - (len>>1); int rlen = len>>1; if(st[rt].lazy3) { ll c = st[rt].lazy3%mdd; st[rt<<1].lazy3 = st[rt<<1|1].lazy3 = st[rt].lazy3; st[rt<<1].lazy1 = st[rt<<1|1].lazy1 = 0; st[rt<<1].lazy2 = st[rt<<1|1].lazy2 = 1; st[rt<<1].sum1 = (llen * (c%mdd))%mdd; st[rt<<1|1].sum1 = (rlen * c)%mdd; st[rt<<1].sum2 = (llen * (c * c)%mdd ) %mdd; st[rt<<1|1].sum2 = ((rlen * ((c * c)%mdd))%mdd)%mdd; st[rt<<1].sum3 = (llen * ((c * c)%mdd * (c % mdd))%mdd)%mdd; st[rt<<1|1].sum3 = (rlen * ((c * c)%mdd * (c % mdd))%mdd)%mdd; st[rt].lazy3 = 0; } if(st[rt].lazy2 != 1 ) { ll c = st[rt].lazy2 % mdd; st[rt<<1].lazy2 = (st[rt<<1].lazy2 * c) % mdd; st[rt<<1|1].lazy2 = (st[rt<<1|1].lazy2 * c) % mdd; if(st[rt<<1].lazy1){ st[rt<<1].lazy1 = (st[rt<<1].lazy1 * c)%mdd; } if(st[rt<<1|1].lazy1){ st[rt<<1|1].lazy1 = (st[rt<<1|1].lazy1 * c)%mdd; } st[rt<<1].sum1 = (st[rt<<1].sum1 * c)%mdd; st[rt<<1|1].sum1 = (st[rt<<1|1].sum1 * c)%mdd; st[rt<<1].sum2 = ((st[rt<<1].sum2 * c%mdd)*c)%mdd; st[rt<<1|1].sum2 = ((st[rt<<1|1].sum2 * c%mdd)*c)%mdd; st[rt<<1].sum3 = ((((st[rt<<1].sum3 * c)%mdd)*c%mdd)*c)%mdd; st[rt<<1|1].sum3 = ((((st[rt<<1|1].sum3 * c)%mdd)*c%mdd)*c)%mdd; st[rt].lazy2 = 1; } if(st[rt].lazy1) { ll c = st[rt].lazy1; st[rt<<1].lazy1 = (st[rt<<1].lazy1 + c)%mdd; st[rt<<1|1].lazy1 = (st[rt<<1|1].lazy1 +c)%mdd; //注意sum3,sum2,sum1 的顺序; st[rt<<1].sum3 = st[rt<<1].sum3 + ((c%mdd)*c%mdd)*c%mdd*llen + 3*c*((st[rt<<1].sum2 + st[rt<<1].sum1 * c)%mdd); st[rt<<1|1].sum3 = st[rt<<1|1].sum3 + ((c%mdd)*c%mdd)*c%mdd*rlen +3*c*((st[rt<<1|1].sum2 + st[rt<<1|1].sum1 * c)%mdd); st[rt<<1].sum2 = st[rt<<1].sum2 + (((llen *c)%mdd) * c)%mdd + 2*(st[rt<<1].sum1)%mdd*c; st[rt<<1|1].sum2 = st[rt<<1|1].sum2 + (((rlen *c)%mdd) * c)%mdd + 2*(st[rt<<1|1].sum1*c)%mdd; st[rt<<1].sum1 = st[rt<<1].sum1 + (llen*c)%mdd; st[rt<<1|1].sum1 = st[rt<<1|1].sum1 + (rlen*c)%mdd; allmdd(rt<<1);allmdd(rt<<1|1); st[rt].lazy1 = 0; } } void build(int l,int r,int rt) { st[rt].sum1 = st[rt].sum2 = st[rt].sum3 = 0; st[rt].lazy1 = st[rt].lazy3 = 0; st[rt].lazy2 = 1; st[rt].l = l,st[rt].r = r; int mid = (l + r) >> 1; if(l == r)return; build(lson); build(rson); pushup(rt); } void update(int l,int r,int rt,int L,int R,int c,int op) { if(l >= L&&r <= R) { if(op==1) { st[rt].lazy1 += c; st[rt].sum3 = (st[rt].sum3 + ((c*c%mdd)*c)%mdd*(r-l+1)%mdd + (3*c*(st[rt].sum2 + st[rt].sum1*c)%mdd))%mdd; st[rt].sum2 = (st[rt].sum2 + (c*c)%mdd*(r-l+1)%mdd + (2 * st[rt].sum1 * c)%mdd)%mdd; st[rt].sum1 = (st[rt].sum1 + (r-l+1)*c%mdd)%mdd; } else if(op==2) { st[rt].lazy2 = (st[rt].lazy2 * c)%mdd; if(st[rt].lazy1) st[rt].lazy1 = (st[rt].lazy1 * c)%mdd; st[rt].sum1 = (st[rt].sum1 * c %mdd)%mdd; st[rt].sum2 = (st[rt].sum2 * ((c * c) %mdd))%mdd; st[rt].sum3 = (st[rt].sum3 * (((c * c)%mdd * c) % mdd))%mdd; } else if(op==3) { st[rt].lazy3 = c; st[rt].lazy1 = 0; st[rt].lazy2 = 1; st[rt].sum1 = (((r-l+1)*c )% mdd); st[rt].sum2 = (((((r-l+1)* c)%mdd )* c %mdd))%mdd; st[rt].sum3 = ((r-l+1)* (((c * c)%mdd * c) % mdd)); } return; } int mid = (l+r)>>1; pushdown(rt); if(L<=mid)update(lson,L,R,c,op); if(R>mid)update(rson,L,R,c,op); pushup(rt); } ll query(int l,int r,int rt,int L,int R,int c) { if(l>=L&&r<=R) { if(c == 1)return st[rt].sum1%mdd; else if(c == 2)return st[rt].sum2%mdd; else return st[rt].sum3%mdd; } int mid = (l+r)>>1; ll ans = 0; pushdown(rt); if(L<=mid)ans = (ans + query(lson,L,R,c))%mdd; if(R>mid)ans = (ans + query(rson,L,R,c))%mdd; return ans; } int main(){ // freopen("input","r",stdin); int n,m; while(scanf("%d%d", &n, &m)==2&&n+m) { // debug(n); build(1,n,1); while(m--) { int op; int a,b,c; scanf("%d%d%d%d", & op, & a,& b, & c); if(op==4){ printf("%lld\n", query(1,n,1,a,b,c)%mdd); } else { update(1,n,1,a,b,c,op); } } } return 0; }
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