. There are m operations within three kinds as following:• 1 x y: change value ax to y.
• 2 x y: change value bx to y.
• 3 k: ask min{t|k≤S(t)}
标签：sample board first represent contains 数据 memset min 输入
. There are m operations within three kinds as following:
2
4 6
2 4 6 8
1 3 5 7
1 2 3
2 3 3
3 15
1 3 8
3 90
3 66
8 5
2 4 8 3 1 3 6 24
2 2 39 28 85 25 98 35
3 67
3 28
3 73
3 724
3 7775
17
87
65
72
58
74
310
2875
还是菜，训练的时候只想到二分，把a相同的项合并到一起来优化时间复杂度。但是没想出来怎么解决合并后向下取整的问题。
$\frac{t-b_i}{a_i}$中令$t=k_1*a_i+c_1$,$b_i=k_2*a_i+c_2$,则$\frac{t-b_i}{a_i} = \frac{k_1*a_i+c_1-k_2*a_i-c_2}{a_i}$
我们只需记录所有$k_2_i$的和，然后t对于每一个ai计算$c_1<c_2$的情况(记为s），$\sum_{i=1}^{1000} k_1_i + \sum_{i=1}^{1000} k_2_i - \sum_{i=1}^{1000} s_i$即为S(t)的值
#include "bits/stdc++.h" using namespace std; typedef long long ll; const int maxn = 1e6; int a[maxn], b[maxn]; int f[1100][1100]; bool check(ll t, ll s) { ll ret = 0; for (int i = 1; i <= 1000; i++) { ret += t / i * f[i][0]; ret -= f[i][t % i + 1]; } return ret >= s; } int main() { // freopen("input.txt", "r", stdin); int _, n, m; scanf("%d", &_); while (_--) { ll ret = 0; scanf("%d %d", &n, &m); memset(f, 0, sizeof(f)); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); for (int i = 1; i <= n; i++) { scanf("%d", &b[i]); ret += b[i] / a[i]; f[a[i]][b[i] % a[i]]++; } for (int i = 1; i <= 1000; i++) { for (int j = i - 1; j >= 0; j--) { f[i][j] += f[i][j + 1]; } } int swi, x, y; while (m--) { scanf("%d %d", &swi, &x); if (swi == 1) { scanf("%d", &y); ret -= b[x] / a[x]; ret += b[x] / y; for (int i = b[x] % a[x]; i >= 0; i--) f[a[x]][i]--; for (int i = b[x] % y; i >= 0; i--) f[y][i]++; a[x] = y; } else if (swi == 2) { scanf("%d", &y); ret -= b[x] / a[x]; ret += y / a[x]; for (int i = b[x] % a[x]; i >= 0; i--) f[a[x]][i]--; for (int i = y % a[x]; i >= 0; i--) { f[a[x]][i]++; } b[x] = y; } else { ll l = 1, r = 1e13, mid; while (l < r) { mid = (l + r) >> 1; if (check(mid, ret + x)) r = mid; else l = mid + 1; } printf("%lld\n", l); } } } return 0; }
标签：sample board first represent contains 数据 memset min 输入
原文地址：https://www.cnblogs.com/albert-biu/p/10661069.html
