初看起来感觉不是很好搞，主要是有赋值操作，我们需要知道的是最近一次在这个行上的赋值操作以及之间的贡献
那么我们离线处理，每个3操作都往前找一个最近的同行2操作，然后两个做差就能得到中间的和。
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
#define A puts("YES")
#define B puts("NO")
using namespace std;
typedef double db;
typedef long long ll;
const ll mo=998244353;
const int N = 2e5 + 5;
struct node {
int op, l, r, x;
};
node a[N];
int n, m, Q, x, r,id;
struct key {
int id, x;
};
vector<key> q[N];
vector<key> v[N];
ll c[N], ans[N];
int lowbit(int x) {
return x & (-x);
}
void upd(int x, ll y) {
for (;x < N;x += lowbit(x)) c[x] += y;
}
ll ask(int x) {
ll s = 0;
for (;x;x -= lowbit(x)) s += c[x];
return s;
}
int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
scanf("%d %d %d", &n, &m, &Q);
fo(i, 1, Q) {
scanf("%d %d %d", &a[i].op, &a[i].l, &a[i].r);
if (a[i].op == 1) scanf("%d", &a[i].x);
}
fd(i, Q, 1) {
if (a[i].op == 1) continue;
if (a[i].op == 2) {
r = a[i].l;
for (int j = 0;j < (int)v[r].size();j++) {
q[i].emplace_back(v[r][j]);
}
v[r].clear();
}
if (a[i].op == 3) {
v[a[i].l].emplace_back((key) { i, a[i].r });
}
}
fo(i, 1, Q) {
if (a[i].op == 1) {
upd(a[i].l, (ll)a[i].x);
upd(a[i].r + 1, (ll)-a[i].x);
}
if (a[i].op == 2) {
for (int j = 0;j < q[i].size();j++) {
x = q[i][j].x;
id = q[i][j].id;
ans[id] = (ll)a[i].r - ask(x);
// printf("%d %d %lld\n", id,a[i].r, ask(x));
}
}
if (a[i].op == 3) {
ans[i] += ask(a[i].r);
}
}
// return 0;
fo(i, 1, Q) {
if (a[i].op == 3) printf("%lld\n", ans[i]);
}
return 0;
}