with ttr for each const -c 描述 master mst boa

時間限制: 10 Sec 內存限制: 128 MB
提交: 98 解決: 32
[提交] [狀態] [命題人:admin]

題目描述

There are two sequences a1,a2,...,an , b1,b2,...,bn . Let . 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)}

輸入

The first line contains a integer T (1≤T≤5) representing the number of test cases.
For each test case, the first line contains two integers n (1≤n≤100000), m (1≤m≤10000).
The following line contains n integers representing the sequence a1,a2,...,an .
The following line contains n integers representing the sequence b1,b2,...,bn .
The following m lines, each line contains two or three integers representing an operation mentioned above.
It is guaranteed that, at any time, we have 1≤ai≤1000, 1≤bi,k≤109 . And the number of queries (type 3 operation) in each test case will not exceed 1000.

輸出

For each query operation (type 3 operation), print the answer in one line.

樣例輸入

復制樣例數據

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
 

樣例輸出

17
87
65
72
58
74
310
2875 

還是菜，訓練的時候只想到二分，把a相同的項合並到一起來優化時間復雜度。但是沒想出來怎麽解決合並後向下取整的問題。

$\frac{t-b_i}{a_i}$中令$t=k_1*a_i+c_1$,$b_i=k_2*a_i+c_2$,則$\frac{t-b_i}{a_i} = \frac{k_1*a_i+c_1-k_2*a_i-c_2}{a_i}$

我們只需記錄所有$k_2$的和，然後t對於每一個ai計算$c_1<c_2$的情況(記為s），$\sum_{i=1}^{1000} k_{1_i} + \sum_{i=1}^{1000} k_{2_i} - \sum_{i=1}^{1000} s_i$即為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;
}

HDU6274 Master of Sequence