（待改）HDU 4578 Transformation （线段树+成段更新+多种操作）

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

题解：线段树+成段更新+多种操作

wa了一天，实在找不到哪里有问题，1次方答案是对的，但往上就错了。
等刷完成段更新的其他题再看看吧。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
//成段更新
const int mod = 10007;
const int MAXN = 1e5 + 5;
int sum[MAXN << 2][4], add[MAXN << 2], mult[MAXN << 2], cover[MAXN << 2];
struct Node {
    int l, r;
    int mid() { return (l + r) >> 1; }
} tree[MAXN << 2];
void fadd(int rt, int c, int m) {
    add[rt] = (add[rt] + c) % mod;
    sum[rt][3] = (sum[rt][3] + (3 * c * sum[rt][2]) % mod + (3 * ((c * c) % mod) * sum[rt][1]) % mod + (((((c * c) % mod) * c) % mod) * m) % mod) % mod;
    sum[rt][2] = (sum[rt][2] + (2 * c * sum[rt][1]) % mod + (((c * c) % mod) * m) % mod) % mod;
    sum[rt][1] = (sum[rt][1] + (c * m) % mod) % mod;
    //int sum1 = sum[rt][1], sum2 = sum[rt][2], sum3 = sum[rt][3];

    //sum[rt][1] += (c * m) % mod;
    //sum[rt][1] %= mod;

    //int tmp2 = (c * c) % mod;
    //sum[rt][2] += (2 * sum1 * c) % mod + (tmp2 * m) % mod;
    //sum[rt][2] %= mod;

    //int tmp3 = (tmp2 * c) % mod;
    //sum[rt][3] += (3 * tmp2 * sum1) % mod + (3 * c * sum2) % mod + (tmp3 * m) % mod;
    //sum[rt][3] %= mod;
}
void fmult(int rt, int c, int m) {
    if (mult[rt]) mult[rt] = (mult[rt] * c) % mod;
    else mult[rt] = c;
    add[rt] = (add[rt] * c) % mod;
    sum[rt][1] = (c * sum[rt][1]) % mod;
    sum[rt][2] = (((c * c) % mod) * sum[rt][2]) % mod;
    sum[rt][3] = (((((c * c) % mod) * c) % mod) * sum[rt][3]) % mod;
    //add[rt] = (add[rt] * c) % mod;

    //int sum1 = sum[rt][1], sum2 = sum[rt][2], sum3 = sum[rt][3];

    //sum[rt][1] = (sum1 * c) % mod;

    //int tmp2 = (c * c) % mod;
    //sum[rt][2] = (sum2 * tmp2) % mod;

    //int tmp3 = (tmp2 * c) % mod;
    //sum[rt][3] = (sum3 * tmp3) % mod;
}
void fcover(int rt, int c, int m) {
    cover[rt] = c;
    add[rt] = mult[rt] = 0;
    sum[rt][1] = (c * m) % mod;
    sum[rt][2] = (c * sum[rt][1]) % mod;
    sum[rt][3] = (c * sum[rt][2]) % mod;
}
void PushDown(int rt, int m) {
    if (cover[rt]) {
        fcover(rt << 1, cover[rt], m - (m >> 1));
        fcover(rt << 1 | 1, cover[rt], m >> 1);
        cover[rt] = 0;
    }
    if (mult[rt]) {
        fmult(rt << 1, mult[rt], m - (m >> 1));
        fmult(rt << 1 | 1, mult[rt], m >> 1);
        mult[rt] = 0;
    }
    if (add[rt]) {
        fadd(rt << 1, add[rt], m - (m >> 1));
        fadd(rt << 1 | 1, add[rt], m >> 1);
        add[rt] = 0;
    }
}
void PushUp(int rt) {
    sum[rt][1] = (sum[rt << 1][1] + sum[rt << 1 | 1][1]) % mod;
    sum[rt][2] = (sum[rt << 1][2] + sum[rt << 1 | 1][2]) % mod;
    sum[rt][3] = (sum[rt << 1][3] + sum[rt << 1 | 1][3]) % mod;
}
void build(int l, int r, int rt) {
    tree[rt].l = l;
    tree[rt].r = r;
    sum[rt][1] = sum[rt][2] = sum[rt][3] = add[rt] = mult[rt] = cover[rt] = 0;
    if (l == r) {
        sum[rt][1] = sum[rt][2] = sum[rt][3] = 0;
        return;
    }
    int m = tree[rt].mid();
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    //PushUp(rt);
}
void update(int c, int l, int r, int rt, int id) {
    if (tree[rt].l == l && r == tree[rt].r) {
        if (id == 1) fadd(rt, c, r - l + 1);
        else if (id == 2) fmult(rt, c, r - l + 1);
        else fcover(rt, c, r - l + 1);
        return;
    }
    if (tree[rt].l == tree[rt].r) return;
    PushDown(rt, tree[rt].r - tree[rt].l + 1);
    int m = tree[rt].mid();
    if (r <= m) update(c, l, r, rt << 1, id);
    else if (l > m) update(c, l, r, rt << 1 | 1, id);
    else {
        update(c, l, m, rt << 1, id);
        update(c, m + 1, r, rt << 1 | 1, id);
    }
    PushUp(rt);
}
int query(int l, int r, int rt, int p) {
    if (l == tree[rt].l && r == tree[rt].r) {
        return sum[rt][p];
    }
    PushDown(rt, tree[rt].r - tree[rt].l + 1);
    int m = tree[rt].mid();
    int res = 0;
    if (r <= m) res += query(l, r, rt << 1, p), res %= mod;
    else if (l > m) res += query(l, r, rt << 1 | 1, p), res %= mod;
    else {
        res += query(l, m, rt << 1, p), res %= mod;
        res += query(m + 1, r, rt << 1 | 1, p), res %= mod;
    }
    return res;
}
int n, m, id, x, y, c;
int main() {
    freopen("in.txt", "r", stdin);
    while (~scanf("%d%d", &n, &m) && n) {
        build(1, n, 1);
        for (int i = 1; i <= m; i++) {
            scanf("%d%d%d%d", &id, &x, &y, &c);
            if (id <= 3) update(c, x, y, 1, id);
            else printf("%d\n", query(x, y, 1, c));
        }
    }
	return 0;
}