이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define int long long 
using ll = long long;
#define fi first
#define si second
#define ar array
#define pb push_back
typedef pair<ll,int> pi;
typedef tuple<int,int,int> ti;  
typedef vector<int> vi;
template<typename T> bool chmin(T &a, T b){return (b < a) ? a = b, 1 : 0;}
template<typename T> bool chmax(T &a, T b){return (b > a) ? a = b, 1 : 0;}
void debug_out() {cerr<<endl;}
template <typename Head, typename... Tail>
void debug_out(Head _H, Tail... _T) {cerr<<" "<<to_string(_H);debug_out(_T...);}
#define debug(...) cerr<<"["<<#__VA_ARGS__<<"]:",debug_out(__VA_ARGS__)
const int N = 100010;
int n;
int h[N], k[N];
int ans[N];
ll total, ret;
multiset<pi> st;
ll fw[N], fw2[N];
void update(int x, int y, ll v) { 
    for (int tx=x; tx < N; tx += tx&(-tx)) fw[tx] += v, fw2[tx] -= v*(x-1);
    for (int ty=y+1; ty < N; ty += ty&(-ty)) fw[ty] -= v, fw2[ty] += v*y; 
}
ll sum (int x) {
    ll res = 0;
    for (int tx=x; tx; tx -= tx&(-tx)) res += fw[tx]*x + fw2[tx];
    return res;
}
ll query(int x, int y) { 
    return sum(y)-sum(x-1);
}
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    cin.exceptions(ios::badbit|ios::failbit);
    cin >> n;
    int mx = 0;
    for (int i = 0; i < n; ++i) {
        cin >> h[i] >> k[i];
        total += k[i];
        chmax(mx, h[i]);
    }
    for (int i = 0; i < n; ++i) update(1, h[i], 1);
    for (int i = 1; i <= mx; ++i) st.insert({query(i, i), i});
    int cur = 0;
    while (st.size()) {
        int rows = st.size();
        if (total >= rows) {
            total -= rows;
            ++cur;
        } else {
            int df = rows - total;
            while (df) {
                ans[st.begin()->si] = cur;
                st.erase(st.find(*st.begin()));
                --df;
            }
            for (auto i: st) ans[i.si] = cur + 1;
            break;
        }   
        while (st.size() && st.begin()->fi == cur) {
            ans[st.begin()->si] = cur;
            st.erase(st.find(*st.begin()));
        }
    }
    for (int i = 1; i <= mx; ++i) {
        ll val = (ll)ans[i] * (ll)(ans[i] + 1) / 2 - ans[i];
        ret += val;
    }
    cout << ret;
    return 0;
}
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