hdu4578（线段树多种区间操作）

/*
translation:
对于一个区间有4个操作：
1.将a~b都加上c
2.将a~b都乘上c
3.将a~b都变成c
4.查询a~b的每个数的p次方的和。（p=1,2,3）
solution:
典型的线段树应用，不过增加了多种操作。一开始没写对，WA了好久，后来意识到更新的时候各个sum之间更新顺序
是有讲究的，于是改了过来，然而还是WA。最后憋得没办法，看题解才发现各个懒惰标记之间往下推的时候也是能够
互相影响的ORZ。这道题思路简单，但是挺复杂的，很考验编码能力。
note:
# 当线段树有多个懒惰标记时，一定要考虑到懒惰标记之间的互相影响。
date:
2016.12.4
*/
#include <iostream>
#include <cstdio>

using namespace std;
const int maxn = 100000 + 5;
const int M = 10007;

typedef long long ll;

int n, q;
int s[maxn*4], e[maxn*4];
ll sum[maxn*4][3], lazy[maxn*4][3];	//0,1,2分别对应于操作1,2,3

void build(int l, int r, int rt)
{
s[rt] = l;
e[rt] = r;
sum[rt][0] = sum[rt][1] = sum[rt][2] = 0;
lazy[rt][0] = lazy[rt][2] = 0;
lazy[rt][1] = 1;

if(l != r){
int m = (l + r) >> 1;
build(l, m, rt << 1);
build(m + 1, r, rt << 1 | 1);
}
}

void push_down(int rt, int len)
{
int lc = rt << 1, rc = rt << 1 | 1;
if(lazy[rt][2]){
lazy[lc][2] = lazy[rc][2] = lazy[rt][2];
lazy[lc][1] = lazy[rc][1] = 1;
lazy[lc][0] = lazy[rc][0] = 0;

ll tmp = (lazy[rt][2] * lazy[rt][2] % M) * lazy[rt][2] % M;
sum[lc][2] = tmp * (len - (len >> 1)) % M;
sum[rc][2] = tmp * (len >> 1) % M;

sum[lc][1] = (((lazy[rt][2] * lazy[rt][2]) % M) * (len - (len >> 1)) % M) % M;
sum[rc][1] = ((lazy[rt][2] * lazy[rt][2] % M) * (len >> 1) % M) % M;

sum[lc][0] = lazy[rt][2] * (len - (len >> 1)) % M;
sum[rc][0] = lazy[rt][2] * (len >> 1) % M;

lazy[rt][2] = 0;
}

if(lazy[rt][1] != 1){
lazy[lc][1] = lazy[lc][1] * lazy[rt][1] % M;
lazy[rc][1] = lazy[rc][1] * lazy[rt][1] % M;
if(lazy[lc][0])	lazy[lc][0] = lazy[lc][0] * lazy[rt][1] % M;
if(lazy[rc][0])	lazy[rc][0] = lazy[rc][0] * lazy[rt][1] % M;

sum[lc][2] = sum[lc][2] * ((lazy[rt][1] * lazy[rt][1] % M) * lazy[rt][1] % M) % M;
sum[rc][2] = sum[rc][2] * ((lazy[rt][1] * lazy[rt][1] % M) * lazy[rt][1] % M) % M;

sum[lc][1] = sum[lc][1] * (lazy[rt][1] * lazy[rt][1] % M) % M;
sum[rc][1] = sum[rc][1] * (lazy[rt][1] * lazy[rt][1] % M) % M;

sum[lc][0] = sum[lc][0] * lazy[rt][1] % M;
sum[rc][0] = sum[rc][0] * lazy[rt][1] % M;

lazy[rt][1] = 1;
}

if(lazy[rt][0]){
lazy[lc][0] += lazy[rt][0];
lazy[rc][0] += lazy[rt][0];

ll tmp = (lazy[rt][0] * lazy[rt][0] % M) * lazy[rt][0] % M;
sum[lc][2] = (sum[lc][2] + (lazy[rt][0] * (sum[lc][1] + lazy[rt][0] * sum[lc][0] % M) % M * 3) % M
+ (len - (len >> 1)) * tmp % M) % M;
sum[rc][2] = (sum[rc][2] + (lazy[rt][0] * (sum[rc][1] + lazy[rt][0] * sum[rc][0] % M) % M * 3) % M
+ (len >> 1) * tmp % M) % M;

tmp = lazy[rt][0] * lazy[rt][0] % M;
sum[lc][1] = (sum[lc][1] + 2 * (lazy[rt][0] * sum[lc][0] % M) % M +
(len - (len >> 1)) * tmp % M) % M;
sum[rc][1] = (sum[rc][1] + 2 * (lazy[rt][0] * sum[rc][0] % M) % M + (len >> 1) * tmp % M) % M;

sum[lc][0] = (sum[lc][0] + (lazy[rt][0] * (len - (len >> 1)) % M)) % M;
sum[rc][0] = (sum[rc][0] + (lazy[rt][0] * (len >> 1) % M)) % M;

lazy[rt][0] = 0;
}
}

void push_up(int rt)
{
int lc = rt << 1, rc = rt << 1 | 1;
sum[rt][0] = (sum[lc][0] + sum[rc][0]) % M;
sum[rt][1] = (sum[lc][1] + sum[rc][1]) % M;
sum[rt][2] = (sum[lc][2] + sum[rc][2]) % M;
}

void update(int t, int c, int st, int ed, int rt)
{
int len = ed - st + 1;
if(s[rt] == st && e[rt] == ed){
if(t == 2){
lazy[rt][2] = c;
lazy[rt][1] = 1;
lazy[rt][0] = 0;
sum[rt][0] = (c*(ed-st+1))%M;
sum[rt][1] = ((c * c % M) * (ed - st + 1)) % M;
sum[rt][2] = (((c * c) % M) * c % M) * (ed - st + 1) % M;
}
else if(t == 1){
lazy[rt][1] = lazy[rt][1] * c % M;
if(lazy[rt][0])	lazy[rt][0] = lazy[rt][0] * c % M;
sum[rt][0] = sum[rt][0] * c % M;
sum[rt][1] = (c * c % M) * sum[rt][1] % M;
sum[rt][2] = ((c * c % M) * c % M) * sum[rt][2] % M;
}
else if(t == 0){
lazy[rt][0] = (lazy[rt][0] + c) % M;
sum[rt][2] = (sum[rt][2] + (sum[rt][1] + c * sum[rt][0] % M) * 3 * c % M +
((((len * c) % M) * c % M) * c % M)) % M;
sum[rt][1] = (sum[rt][1] + (sum[rt][0] * c % M) * 2 % M + (len * c % M) * c % M) % M;
sum[rt][0] = (sum[rt][0] + c * len % M) % M;
}
return;
}
if(s[rt] == e[rt])	return;

push_down(rt, e[rt] - s[rt] + 1);

int m = (s[rt] + e[rt]) >> 1;
if(ed <= m)	update(t, c, st, ed, rt << 1);
else if(st > m)	update(t, c, st, ed, rt << 1 | 1);
else{
update(t, c, st, m, rt << 1);
update(t, c, m+1, ed, rt << 1 | 1);
}
push_up(rt);
}

ll query(int p, int l, int r, int rt)
{
if(s[rt] == l && e[rt] == r){
return sum[rt][p] % M;
}

push_down(rt, e[rt] - s[rt] + 1);

int m = (s[rt] + e[rt]) >> 1;
ll res = 0;

if(r <= m)	res += query(p, l, r, rt << 1) % M;
else if(l > m)	res += query(p, l ,r, rt << 1 | 1) % M;
else{
res += query(p, l, m, rt << 1) % M;
res += query(p, m+1, r, rt << 1 | 1) % M;
}
return res % M;
}

int main()
{
//freopen("in.txt", "r", stdin);
while(~scanf("%d%d", &n, &q)){
if(n == 0 && q == 0)	break;

build(1, n, 1);

while(q--){
int op, a, b, c;
scanf("%d%d%d%d", &op, &a, &b, &c);

if(op == 4){
printf("%lld\n", query(c-1, a, b, 1) % M);
}else{
update(op-1, c, a, b, 1);
}
}
}
}