线段树 HDU - 4578 计算平方和 立方和
https://vjudge.net/problem/HDU-4578
参考自https://blog.csdn.net/shiqi_614/article/details/9939485
题意:给你一个数组,初始值为零,有四种操作
(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
将每个节点看作ax+b
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。
设操作1、2、3的懒操作标记分别是add,mult,same。当三个值同时存在时,add和mult一定是后来到的(因为same会把add和mult清空),所以我们先传递的是same。对于后两个懒标记,我的处理方法是,再者传递mult,最后add。因为当把一个区间标记上mult时,如果发现这个区间的add不为0,那么我会把add更新为add*mult。
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <string> #include <set> #include <map> #include <stack> #include <cmath> #include <algorithm> #include <queue> #define INF (1<<30) using namespace std; #define LL(x) (x<<1) #define RR(x) (x<<1|1) #define MID(a,b) (a+((b-a)>>1)) const int MOD = 10007; const int N = 1e5 + 5; int sum1, sum2, sum3, tmp2, tmp3; int add[N * 4], mult[N * 4], same[N * 4], sum[N * 4][4], sz[N * 4]; void fun_add(int ind, int valu){ add[ind] = (add[ind] + valu) % MOD; sum1 = sum[ind][1], sum2 = sum[ind][2], sum3 = sum[ind][3]; sum[ind][1] += (valu * sz[ind]) % MOD; sum[ind][1] %= MOD; tmp2 = (valu * valu) % MOD; sum[ind][2] += (2 * sum1 * valu) % MOD + (tmp2 * sz[ind]) % MOD; sum[ind][2] %= MOD; tmp3 = (tmp2 * valu) % MOD; sum[ind][3] += (3 * tmp2 * sum1) % MOD + (3 * valu * sum2) % MOD + (tmp3 * sz[ind]) % MOD; sum[ind][3] %= MOD; } void fun_mult(int ind, int valu){ if (mult[ind]) mult[ind] = (mult[ind] * valu) % MOD; else mult[ind] = valu; add[ind] = (add[ind] * valu) % MOD; sum1 = sum[ind][1], sum2 = sum[ind][2], sum3 = sum[ind][3]; sum[ind][1] = (sum1 * valu) % MOD; tmp2 = (valu * valu) % MOD; sum[ind][2] = (sum2 * tmp2) % MOD; tmp3 = (tmp2 * valu) % MOD; sum[ind][3] = (sum3 * tmp3) % MOD; } void fun_same(int ind, int valu){ same[ind] = valu; add[ind] = mult[ind] = 0; sum[ind][1] = (valu * sz[ind]) % MOD; tmp2 = (valu * valu) % MOD; sum[ind][2] = (tmp2 * sz[ind]) % MOD; tmp3 = (tmp2 * valu) % MOD; sum[ind][3] = (tmp3 * sz[ind]) % MOD; } void PushUp(int ind){ sum[ind][1] = (sum[LL(ind)][1] + sum[RR(ind)][1]) % MOD; sum[ind][2] = (sum[LL(ind)][2] + sum[RR(ind)][2]) % MOD; sum[ind][3] = (sum[LL(ind)][3] + sum[RR(ind)][3]) % MOD; } void PushDown(int ind){ if (same[ind]){ fun_same(LL(ind), same[ind]); fun_same(RR(ind), same[ind]); same[ind] = 0; } if (mult[ind]){ fun_mult(LL(ind), mult[ind]); fun_mult(RR(ind), mult[ind]); mult[ind] = 0; } if (add[ind]){ fun_add(LL(ind), add[ind]); fun_add(RR(ind), add[ind]); add[ind] = 0; } } void build(int lft, int rht, int ind){ add[ind] = mult[ind] = same[ind] = 0; sum[ind][1] = sum[ind][2] = sum[ind][3] = 0; sz[ind] = rht - lft + 1; if (lft != rht){ int mid = MID(lft, rht); build(lft, mid, LL(ind)); build(mid + 1, rht, RR(ind)); } } void updata(int st, int ed, int valu, int type, int lft, int rht, int ind){ if (st <= lft && rht <= ed){ if (type == 1) fun_add(ind, valu); else if (type == 2) fun_mult(ind, valu); else fun_same(ind, valu); } else{ PushDown(ind); int mid = MID(lft, rht); if (st <= mid) updata(st, ed, valu, type, lft, mid, LL(ind)); if (ed > mid) updata(st, ed, valu, type, mid + 1, rht, RR(ind)); PushUp(ind); } } int query(int st, int ed, int p, int lft, int rht, int ind){ if (st <= lft && rht <= ed) return sum[ind][p]; else{ PushDown(ind); int mid = MID(lft, rht), sum1 = 0, sum2 = 0; if (st <= mid) sum1 = query(st, ed, p, lft, mid, LL(ind)); if (ed > mid) sum2 = query(st, ed, p, mid + 1, rht, RR(ind)); PushUp(ind); return (sum1 + sum2) % MOD; } } int main(){ int n, m; while (scanf("%d%d", &n, &m) != EOF){ if (n == 0 && m == 0) break; int type, x, y, c; build(1, n, 1); for (int i = 1; i <= m; i++){ scanf("%d%d%d%d", &type, &x, &y, &c); if (type != 4) updata(x, y, c, type, 1, n, 1); else printf("%d\n", query(x, y, c, 1, n, 1)); } } return 0; }