Codeforces Round #299 (Div. 2) C. Tavas and Karafs 二分搜索+数列

C. Tavas and Karafs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.

Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs issi = A + (i - 1) × B.

For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.

Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find the largest number r such thatl ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.

Input

The first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).

Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.

Output

For each query, print its answer in a single line.

Examples

input

2 1 41 5 33 3 107 10 26 4 8

output

4-18-1

input

1 5 21 5 102 7 4

output

12

Source

Codeforces Round #299 (Div. 2)

My Solution

题意：每个萝卜长度为 hi = a + (i - 1) * b，然后每次询问是每次操作最多把 m 个不同的未吃完的萝卜每个咬掉1单位长度，最多 t 次操作，其中求最大的r，是的[l......r]访问内的萝卜被这 t，m的操作吃完

二分搜索+数列

首先r必须且只要满足2个条件即可，1）a + (r - 1) * b <= t

                                                                2) sum{hi | l <= i <= r } <= t * m

所以对于每个询问进行一次二分搜索即可。

suml = (2 * a + ((l - 1) - 1) * b) * (l - 1) / 2;
LL x = l, y = (t - a) / b + 1 + 1, mid;    // y = (t - a) / b + 1 + 1 多加个1不上整数除法的精度损失， 也可以直接用y = 2e6
while(x + 1 < y){
    mid = (x + y) >> 1;
    if(check(a, b, suml, t, m, mid)) x = mid;
    else y = mid;

}

然后答案可能在x里也可能在y里，且y比x更优，所以

if(check(a, b, suml, t, m, y)) r = y;
else if(check(a, b, suml, t, m, x)) r = x;
else r = -1;

即可

复杂度 O(nlogn)

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;const int maxn = 1e6 + 8;inline bool check(const LL &a, const LL &b, LL &suml, const LL &t, const LL &m, const LL &mid){    if(a + (mid - 1) * b > t) return false;    if((2*a + (mid - 1) * b) * mid / 2 - suml > t * m) return false;    else return true;}int main(){    #ifdef LOCAL    freopen("c.txt", "r", stdin);    //freopen("c.out", "w", stdout);    int T = 2;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    LL a, b, n, l, t, m, r, suml;    cin >> a >> b >> n;    while(n--){        r = -1;        cin >> l >> t >> m;        suml = (2 * a + ((l - 1) - 1) * b) * (l - 1) / 2;        LL x = l, y = (t - a) / b + 1 + 1, mid;    // y = (t - a) / b + 1 + 1 多加个1不上整数除法的精度损失， 也可以直接用y = 2e6        while(x + 1 < y){            mid = (x + y) >> 1;            if(check(a, b, suml, t, m, mid)) x = mid;            else y = mid;        }        //cout << x << " " << y << endl;        if(check(a, b, suml, t, m, y)) r = y;        else if(check(a, b, suml, t, m, x)) r = x;        else r = -1;        if(n) cout << r << "\n";        else cout << r << endl;    }    #ifdef LOCAL    cout << endl;    }    #endif // LOCAL    return 0;}

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