P3373 线段树 2

考点

  • 线段树

题解

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 50;
int n, m, mod, a[maxn];

struct {
int l, r;
ll s, tag_add, tag_mul;
#define l(x) tr[x].l
#define r(x) tr[x].r
#define s(x) tr[x].s
#define tag_add(x) tr[x].tag_add
#define tag_mul(x) tr[x].tag_mul
#define lc(x) x << 1
#define rc(x) (x << 1) + 1
} tr[4 * maxn];

void up(int p) { s(p) = (s(lc(p)) + s(rc(p))) % mod; }

void build(int p, int l, int r) {
l(p) = l, r(p) = r;
s(p) = 0, tag_add(p) = 0, tag_mul(p) = 1;
if (l == r) {
s(p) = a[l];
return;
}
int mid = (l + r) / 2;
build(lc(p), l, mid), build(rc(p), mid + 1, r);
up(p);
}

void add(int p, ll v) {
s(p) = (s(p) + (r(p) - l(p) + 1) * v) % mod;
tag_add(p) = (tag_add(p) + v) % mod;
}

void mul(int p, ll v) {
s(p) = (s(p) * v) % mod;
tag_mul(p) = (tag_mul(p) * v) % mod;
tag_add(p) = (tag_add(p) * v) % mod;
}

void down(int p) {
if (tag_mul(p) != 1) {
mul(lc(p), tag_mul(p)), mul(rc(p), tag_mul(p));
tag_mul(p) = 1;
}
if (tag_add(p)) {
add(lc(p), tag_add(p)), add(rc(p), tag_add(p));
tag_add(p) = 0;
}
}

void update(int p, int l, int r, ll v, int opt) {
if (l <= l(p) && r(p) <= r) {
if (opt == 1) mul(p, v);
if (opt == 2) add(p, v);
return;
}
down(p);
int mid = (l(p) + r(p)) / 2;
if (l <= mid) update(lc(p), l, r, v, opt);
if (r > mid) update(rc(p), l, r, v, opt);
up(p);
}

ll ask(int p, int l, int r) {
if (l <= l(p) && r(p) <= r) return s(p);
down(p);
int mid = (l(p) + r(p)) / 2;
ll ans = 0;
if (l <= mid) ans = (ans + ask(lc(p), l, r)) % mod;
if (r > mid) ans = (ans + ask(rc(p), l, r)) % mod;
return ans;
}

int main() {
scanf("%d%d%d", &n, &m, &mod);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
build(1, 1, n);
int opt, x, y;
ll k;
while (m--) {
scanf("%d", &opt);
if (opt == 1 || opt == 2) {
scanf("%d%d%lld", &x, &y, &k);
update(1, x, y, k, opt);
} else {
scanf("%d%d", &x, &y);
printf("%lld\n", ask(1, x, y));
}
}
return 0;
}

思路

增加的操作,相当于 \[ s\prime=s+nk \\ s\prime\text{新区间和,}s\text{旧区间和,}n\text{区间个数,}k\text{单位变化量} \] 举个例子,有区间{1, 2, 3},原区间和等于6

新增k等于2,相等于新s = 6 + 3 * 2 = 12

乘法的操作,相当于 \[ s\prime=sk \\ s\prime\text{新区间和,}s\text{旧区间和,}k\text{单位变化量} \] 还是上述那个例子,新乘k等于2,相等于新s = 6 * 2 = 12


所以可以模拟以下流程,假设原和为s

  1. k1 \[ s\prime=s+nk_1 \]

  2. k2 \[ s\prime=sk_2+nk_1k_2 \]

  3. k3 \[ s\prime=sk_2+n\left( k_1k_2+k_3 \right) \]

  4. k4 \[ s\prime=sk_2k_4+n\left( k_1k_2+k_3 \right) k_4 \]

根据上述规律可以得到抽象式子,新s = s * tag_mul + n * 单位变化量

单位变化量同时被tag_add累加、tag_mul累乘影响。

所以每次如果有新的乘法,除了累乘到tag_mul之外,还要同时tag_add执行乘法