这道题放在了T1.本来以为挺简单的，结果愣是卡了我1小时，还没做出来，先ko了T3，才把这道题的暴力写了写，拿了40pts
这道题总的来说思路挺简单的，就是没想出来，看思路吧：
我们能注意到，给的两类操作并不会改变单调性：队以仁义的x≤y，在操作后仍然满足x`≤y`
我们将原序列升序排列，可以分别二分出最大和最小的下标
时间复杂度O(n logmax|Ai| )
为了防止超时可以开__int128
下面是暴力代码（40pts）
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5 + 10;
long long read() {long long x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') f = -1;ch = getchar();}while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();return x * f;
}
int n, q, l, r, ans;
int a[N];
signed main() {freopen("arithmetic.in", "r", stdin);freopen("arithmetic.out", "w", stdout);n = read(), q = read(), l = read(), r = read();for (int i = 1; i <= n; i++) {a[i] = read();}while (q--) {int opt, x, s, t;opt = read(), x = read(), s = read(), t = read();if (opt == 1) {for (int i = 1; i <= n; i++) {if (a[i] >= x) {a[i] = (a[i] + s) * t;}}} else {for (int i = 1; i <= n; i++) {if (a[i] <= x) {a[i] = (a[i] - s) / t;}}}}for (int i = 1; i <= n; i++) {if (a[i] >= l && a[i] <= r) ans++;}printf("%lld", ans);return 0;
}AC代码：
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef __int128 i128;
int n, q, qry[200010][4], a[200010], L, R;
i128 foo(int x) {i128 ans = x;for (int i = 1; i <= q; i++) {if (qry[i][0] == 1 && ans >= qry[i][1])ans = (ans + qry[i][2]) * qry[i][3];if (qry[i][0] == 2 && ans <= qry[i][1])ans = (ans - qry[i][2]) / qry[i][3];}return ans;
}
signed main() {freopen("arithmetic.in", "r", stdin);freopen("arithmetic.out", "w", stdout);cin.tie(0)->sync_with_stdio(false);cin >> n >> q >> L >> R;for (int i = 1; i <= n; i++) cin >> a[i];sort(a + 1, a + 1 + n);for (int i = 1; i <= q; i++) cin >> qry[i][0] >> qry[i][1] >> qry[i][2] >> qry[i][3];int l = 1, r = n, ans = n + 1;while (l <= r) {int mid = (l + r) >> 1;i128 x = foo(a[mid]);if (L <= x) {ans = mid;r = mid - 1;} elsel = mid + 1;}int sws = 0;l = 1, r = n;while (l <= r) {int mid = (l + r) >> 1;i128 x = foo(a[mid]);if (x <= R) {sws = mid;l = mid + 1;} elser = mid - 1;}if (ans > sws)puts("0");elseprintf("%lld\n", sws - ans + 1);return 0;
}
