hdu 4578 Transformation（线段树中级，区间和加强）

题意：
在区间和基础版本上增加了两个新的东西：
1）将一段区间的数都乘上某个系数。
2）查询区间的平方和，立方和。
思路：
为每个节点设置 add 和 mul 标记。
每个数的实际值就是 x * mul + add，
实际平方和：∑ri=l(ai∗mul+add)2
实际立方和：∑ri=l(ai∗mul+add)3
所以我们对每个节点维护，∑ai，∑ai2和∑ai3这三个和，就可以根据式子求出实际的和了。

#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)+1)int n, m;struct node {    int add, mul, set, sum[3]; // leaf`s real value: sum[0] * mul + add    inline int s(int x) {return sum[x];}    inline int p1(int L, int R) { // get sum of 1-th power        int s1 = s(0), k = R - L + 1;        return (s1*mul%Mod + k*add%Mod) % Mod;    }    inline int p2(int L, int R) { // get sum of 2-th power        int s1 = s(0), s2 = s(1), k = R - L + 1, m2 = mul*mul%Mod, x2 = add*add%Mod;        s2 = s2*m2%Mod, s1 = s1*mul%Mod;        return (s2%Mod + 2*add*s1%Mod+k*x2%Mod) % Mod;    }    inline int p3(int L, int R) { // // get sum of 3-th power        int s1 = s(0), s2 = s(1), s3 = s(2), k = R - L + 1, m2 = mul*mul%Mod,            m3 = m2 * mul % Mod, x2 = add*add%Mod, x3 = x2*add%Mod;        s3 = s3*m3%Mod, s2 = s2*m2%Mod, s1 = s1*mul%Mod;        return (k*x3%Mod + s3 + 3*x2*s1%Mod + 3*add*s2%Mod) % Mod;    }};node a[Maxn*4+5];void mul(int o, int x) {    a[o].mul = a[o].mul * x % Mod; a[o].add = a[o].add * x % Mod;}void add(int o, int x) {    a[o].add = (a[o].add + x) % Mod;}void Set(int L, int R, int o, int x) {    a[o].set = x; a[o].add = 0; a[o].mul = 1;    a[o].sum[0] = x * (R-L+1) % Mod;    a[o].sum[1] = a[o].sum[0] * x % Mod;    a[o].sum[2] = a[o].sum[1] * x % Mod;}void segDown(int L, int R, int o) {   if (L == R) return;        int lc = lson(o), rc = rson(o), mid = (L+R)>>1;        if (a[o].set != -1) {            Set(L, mid, lc, a[o].set); Set(mid+1, R, rc, a[o].set); a[o].set = -1;        }        if (a[o].mul > 1) {            mul(lc, a[o].mul); mul(rc, a[o].mul); a[o].mul = 1;        }        if (a[o].add) {            add(lc, a[o].add); add(rc, a[o].add); a[o].add = 0;        }}void segUp(int L, int R, int o) {    if (L == R) return;    int lc = lson(o), rc = rson(o), mid = (L+R)>>1;    a[o].set = -1; a[o].add = 0; a[o].mul = 1;    a[o].sum[0] = (a[lc].p1(L, mid) + a[rc].p1(mid+1, R)) % Mod;    a[o].sum[1] = (a[lc].p2(L, mid) + a[rc].p2(mid+1, R)) % Mod;    a[o].sum[2] = (a[lc].p3(L, mid) + a[rc].p3(mid+1, R)) % Mod;}void segUpdate(int L, int R, int o, int qL, int qR, int type, int val) {    //cout << "upd: " << L << ' ' << R << ": " << endl;    if (qL <= L && R <= qR) {            if (type == 1) {                add(o, val);            } else if (type == 2) {                mul(o, val);            } else if (type == 3) {                Set(L, R, o, val);            }            return;    }    segDown(L, R, o);    int lc = lson(o), rc = rson(o), mid = (L+R)>>1;    if (qL <= mid) segUpdate(L, mid, lc, qL, qR, type, val);    if (qR > mid) segUpdate(mid+1, R, rc, qL, qR, type, val);    segUp(L, R, o);}int segAsk(int L, int R, int o, int qL, int qR, int p) {    if (qL <= L && R <= qR) {            if (p == 1) return a[o].p1(L, R);            if (p == 2) return a[o].p2(L, R);            if (p == 3) return a[o].p3(L, R);    }    segDown(L, R, o);    int lc = lson(o), rc = rson(o), mid = (L+R)>>1, ret = 0;    if (qL <= mid) ret += segAsk(L, mid, lc, qL, qR, p);    if (qR >  mid) ret += segAsk(mid+1, R, rc, qL, qR, p);    segUp(L, R, o);    return ret % Mod;}void segBuild(int L, int R, int o) {    a[o].set = -1; a[o].add = 0; a[o].mul = 1;    a[o].sum[0] = a[o].sum[1] = a[o].sum[2] = 0;    if (L < R) {        int lc = lson(o), rc = rson(o), mid = (L+R)>>1;        segBuild(L, mid, lc); segBuild(mid+1, R, rc);    }}// debugvoid segPrint(int L, int R, int o) {    if (o == 1) {cout << "----print----" << endl;}    cout << "(" << L << ", " << R << "), " << o << ": " << a[o].add << ", " << a[o].sum[0] <<  ' ' << a[o].sum[1] << ' ' << a[o].sum[2] << endl;    if (L < R) {        int lc = lson(o), rc = rson(o), mid = (L+R)>>1;        segPrint(L, mid, lc); segPrint(mid+1, R, rc);    }}int main() {#ifndef ONLINE_JUDGE    freopen("input.in", "r", stdin);#endif    while (cin >> n >> m && (n+m)) {        segBuild(1, n, 1);        int t, x, y, z;        rep(i, 1, m) {            cin >> t >> x >> y >> z;            if (t != 4) {                segUpdate(1, n, 1, x, y, t, z);            } else {                cout << segAsk(1, n, 1, x, y, z) << endl;            }        }    }    return 0;}