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>
#define int long long
#define ll long long
#define pii pair<int, int>
#define F first
#define S second
using namespace std;
const bool LittleCubeIsBurningChicken = true;
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll N, L, v[2005][4005], sum[4005];
int abtodc(int a, int b, int c)
{
return (b * c + a - 1) / a;
}
vector<pii> sol;
pii solve(vector<int> p)
{
sol.clear();
pii last = {0, 1};
int nxt = 1;
for (auto i : p)
{
ll tmp = sum[i];
while (tmp >= v[i][nxt] - abtodc(last.S, last.F, v[i][nxt]))
tmp -= v[i][nxt] - abtodc(last.S, last.F, v[i][nxt]), last = {0, 1}, nxt++;
last = {abtodc(last.S, last.F, v[i][nxt]), v[i][nxt]};
last.F += tmp;
sol.emplace_back(pii{last.F + last.S * (nxt - 1), last.S});
}
return sol.back();
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cin >> N >> L;
// why divide by N?
// how to solve using only fractions? -> we can discard leftovers (things cannot be converted to current fraction) from last person
// is it guarnteed to always have a solution? -> yes (at least for N <= 6)
// why?
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= L; j++)
{
cin >> v[i][j];
// sum >= sum of v / N
// sum * N >= sum of v
sum[i] += v[i][j];
v[i][j] *= N;
}
for (int j = L + 1; j <= L + N; j++)
v[i][L + 1] = 2'00000'000;
}
vector<int> p;
for (int i = 1; i <= N; i++)
p.emplace_back(i);
solve(p);
while (sol.back().F > sol.back().S * L)
{
pii cur = sol.back();
for (int i = 0; i < N - 1; i++)
{
swap(p[i], p[i + 1]);
solve(p);
if (sol.back().F * cur.S < sol.back().S * cur.F)
break;
swap(p[i], p[i + 1]);
}
}
sol.pop_back();
for (auto [a, b] : sol)
cout << a << ' ' << b << '\n';
for (auto i : p)
cout << i << ' ';
cout << '\n';
}
컴파일 시 표준 에러 (stderr) 메시지
naan.cpp: In function 'int main()':
naan.cpp:79:15: warning: structured bindings only available with '-std=c++17' or '-std=gnu++17'
79 | for (auto [a, b] : sol)
| ^
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