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：$\sum_{i=l}^{r}{(a_i *mul+add)}^2$
实际立方和：$\sum_{i=l}^{r}{(a_i *mul+add)}^3$
所以我们对每个节点维护，$\sum{ai}$，$\sum{ai^2}$和$\sum{ai^3}$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);
    }
}

// debug
void 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;
}