TG D 2

#include <iostream>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <set>
#include <map>
#define pii pair <int, int>
#define pb(x) push_back(x)
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
void cv(vector <int> &v) {
    for (auto x : v) cout << x << ' ';
    cout << "\n";
}

void cvl(vector <ll> &v) {
    for (auto x : v) cout << x << ' ';
    cout << "\n";
}


void cvv(vector <vector <int> > &v) {
    for (auto x : v) cv(x);
    cout << "\n";
}

void cvb(vector <bool> v) {
    for (bool x : v) cout << x << ' ';
    cout << "\n";
}

void cvs(vector <string>  v) {
    for (auto a : v) {
        cout << a << "\n";
    }
}

void cvp(vector <pii> a) {
    for (auto p : a) {
        cout << p.first << ' ' << p.second << "\n";
    }
    cout << "\n";
}

struct prt{
    ll r, b, id;
    void see() {
        cout << r << ' ' << b << ' ' << id << "\n\n";
    }
};

bool cmp(prt a, prt b) {
    return a.r < b.r;
}

struct res{
    ll spent, id, L, rk;//M: M - 1 -- макс кроме нас
    void see() {
        cout << "res\n";
        cout << "spent id M rk\n";
        cout << spent << ' '<< id << ' ' << L << ' ' << rk << "\n\n";
    }
};

bool cmp1(res a, res b) {
    return a.spent < b.spent;
}


bool sh = 1;

/*
6
1 1
1 2
3 3
3 4
5 7
5 8

8
10 10
10 20
30 30
30 40
50 70
50 80
100 200
100 300
*/

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n; cin >> n;
    vector <prt> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i].r >> a[i].b;
        a[i].id = i;
    }
    sort(a.begin(), a.end(), cmp);
    vector <ll> pf(n);
    pf[0] = a[0].r;
    for (int i = 1; i < n; ++i) {
        pf[i] = pf[i - 1] + a[i].r;
    }
    vector <res> ans;

    if (sh) {
        cout << "a\n";
        for (auto y: a) {
            y.see();
        }
        cout << "pf\n";
        cvl(pf);
    }

    for (int i = 0; i < n; ++i) {
        if (a[i].b == -1) {
            continue;
        }
        ll L = a[i].r, R = a[n - 1].r, M, ad;
        if (sh) {
            cout << "i = " << i << "\n";
            cout << "L R = " << L << ' ' << R << "\n";
        }
        int lk, rk, mk;
        while (L + 1 < R) {
            M = (L + R) / 2;
            if (sh) {
                cout << "M = " << M << "\n";
            }
            lk = i - 1;
            rk = n - 1;
            if (sh) {
                cout << "lk rk = " << lk << ' ' << rk << "\n";
            }
            while (lk + 1 < rk) {
                mk = (lk + rk) / 2;
                if (sh) {
                    cout << "mk = " << mk << "\n";
                }
                if (a[mk].r >= M) {
                    rk = mk;
                } else {
                    lk = mk;
                }
            }
            ll del = 0;
            if (rk > 0) {
                del = pf[rk - 1];
            }
            if (sh) {
                cout << "rk del = " << rk << ' ' << del << "\n";
            }
            ad = pf[n - 1] - del - (M - 1) * (n - rk);
            if (sh) {
                cout << "(n - rk) = " << (n - rk ) << "\n";
                cout << "ad (cut voices) = " << ad << "\n";
            }
            if (a[i].r + ad >= M) {
                L = M;
            } else {
                R = M;
            }
        }
        //НЕТ!!! СРАЗУ ТУТ НАДО ДЕЛАТЬ ПРОВЕРКУ НА ОПТИМАЛЬНОСТЬ
        ll spent = L - a[i].r + a[i].b;  //тк нам доостаточно уровня M!!!
        //а зная итоговое rk res и M, мы смодем восстановить, что каким-то партиям мы вернем срезанные голоса
        res now = {spent, a[i].id, L, rk};
        ans.push_back(now);

        if (sh) {

            cout << "now\n";
            now.see();
        }
    }
    sort(ans.begin(), ans.end(), cmp1);

    if (sh) {
        cout << "ans\n";
        for (auto y: ans) {
            y.see();
        }
    }

    if (sh) {
        cout << "RES ANS\n";
        ans[0].see();
    }
}