hdu　6276　Master of Sequence　二分加预处理-优快云博客

本文链接：https://blog.youkuaiyun.com/renzijing/article/details/89060449

问题 K: Master of Sequence

时间限制: 10 Sec 内存限制: 128 MB
提交: 106 解决: 38
[提交] [状态] [命题人:admin]

题目描述

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)}

输入

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.

输出

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

样例输入

复制样例数据

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

样例输出

[提交][状态]

对于原式子 $S(t)=\sum_{i=1}^{n} \left \lfloor \frac{t- b_{i}}{ a_{i}}\right \rfloor$ 　，令 $t=a_{i} * k_{1} + c_{1}$ ， $b_{i} = a_{i} * k _{2} + c_{2}$ ，则原式子可以化为　 $k_{1} - k_{2} + \frac{c_{1} - c_{2}}{a_{i}}$

当　 $c_{1}>= c_{2}$ 　时，值为　 $k_{1} - k_{2}$

当　 $c_{1} < c_{2}$ 时，值为　 $k_{1} - k_{2} -1$ 　，对于这种情况，向下取整会多算１，所以减去这种情况就可以了

那么 $s(t) = \sum_{i=1}^{n} (k_{1} - k_{2} - (c_{1} < c_{2}))$ ，数组b a都是固定的，可以先处理出所有的 $k_{2}$

式子就变成了 $\sum_{i=1}^{n} (k_{1} - (c_{1} < c_{2})) >= k + \sum_{i=1}^{n} k_{2}$ ，式子是单调变化的

二分check时 t 确定，就可以直接计算了

ａ的范围到1000，那么就可以直接预处理

用一个二维数组f[ i ][ j ]　表示当a==i时，b%a >= j的情况有多少种

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
const int inf = 1e9;
const int mod = 998244353;
int a[maxn], b[maxn];
int f[1100][1100];

bool check(ll t, ll x) {
    ll res = 0;
    for (int i = 1; i <= 1000; i++) {
        res += t / i * f[i][0];
        res -= f[i][t % i + 1];  //减去c1<c2的情况
    }
    return res >= x;
}

ll sol(ll x) {
    ll l = 0, r = 1e13;
    ll ans = 0;
    while (l <= r) {
        ll mid = (l + r) >> 1;
        if (check(mid, x)) {
            ans = mid;
            r = mid - 1;
        } else l = mid + 1;
    }
    return ans;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        ll ret = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &b[i]);
            ret += b[i] / a[i]; //ret就是k2
            f[a[i]][b[i] % a[i]]++;
        }
        for (int i = 1; i < 1005; i++) {
            for (int j = i - 1; j >= 0; j--) {
                f[i][j] += f[i][j + 1];
            }
        }
        int p, x, y;
        while (m--) {
            scanf("%d", &p);
            if (p == 1) {
                scanf("%d%d", &x, &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 (p == 2) {
                scanf("%d%d", &x, &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 k;
                scanf("%lld", &k);
                printf("%lld\n", sol(ret + k));
            }
        }
        memset(f, 0, sizeof(f));
    }
    return 0;
}