洛谷__P12000 扶苏出勤日记（单调栈+二分）

题目链接：P12000 扶苏出勤日记 - 洛谷

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题目大意：

每天赚 b[i]​ 块钱，$1$ 块钱可以换 $a[i]$ 个币去打乌蒙，

钱 和 币都可以存，

求每天打乌蒙的最多次数。

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代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#include<array>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/numeric>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define mst(a,x) memset(a,x,sizeof (a))
#define gmap __gnu_pbds::gp_hash_table
#define power __gnu_cxx::power
using namespace std;
typedef pair<int, int> pii;

const int N = 2000086, mod = 998244353;

int n, m;
int a[N], b[N];
int p[N];

bool check(int mid) {

    int moy = 0, hav = 0;//金钱 , 游戏币数
    for (int i = 1; i <= n; i++) {
        moy += b[i];
        if (p[i]) { //要攒钱
            if (hav < (p[i] - i) * mid) {//撑不到下一个更优惠的时间点
                int t = (p[i] - i) * mid - hav;//还需要的游戏币
                int us = min(t / a[i] + (t % a[i] > 0), moy);//花掉的钱
                hav += us * a[i];
                moy -= us;
            }
        } else {
            hav += moy * a[i];
            moy = 0;
        }
        
        hav -= mid;
        if (hav < 0) return false; 
    }
    
    return true; 
}

void solve() {

    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i];
    
    stack<int> st;//单调栈：右边第一个比当前点大的数的位置
    for (int i = n; i; i--) {
        while (st.size() && a[i] >= a[st.top()]) st.pop();
        p[i] = st.empty() ? 0 : st.top();
        st.push(i);
    }
    
    int l = 0, r = 1e12;
    while (l + 1 < r) {
        int mid = l + r >> 1;
        if (check(mid)) l = mid;    // 如果mid可行，尝试更大的值
        else r = mid ;            // 如果mid不可行，尝试更小的值
    }
    cout << l << endl;
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
    cin >> T;
    while (T--) solve();
    
    return 0;
}