oj.uz

답안 #123872

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
123872 2019-07-02T08:23:50 Z 이온조(#3029) Naan (JOI19_naan) C++14
0 / 100
 3 ms 376 KB 
#include <bits/stdc++.h>
using namespace std;
using pll = pair<long long, long long>;
#define X first
#define Y second

const pll INF = {1LL * 1e18, 1};
int N, L, V[2009][2009], A[2009];

bool operator <(pll P, pll Q) {
	return 1LL * P.X * Q.Y < 1LL * Q.X * P.Y;
}

pll f(long long a, long long b) {
	long long gcd = __gcd(a, b);
	a /= gcd; b /= gcd;
	while(b >= 1000000000LL) {
		if(a & 1) ++a;
		if(b & 1) --b;
		a >>= 1LL; b >>= 1LL;
	}
	return {a, b};
}

pll R(int i, pll st) {
	if(st > INF) return INF;
	long long div = st.X / st.Y, mod = st.X % st.Y, lft = st.Y - mod;
	long long a, b;
	// lft / st.Y * V[i][div] >= A[i] / N
	if(1LL * lft * V[i][div] * N >= 1LL * A[i] * st.Y) {
		// ((a / b) - (st.X / st.Y)) * V[i][div] == A[i] / N
		// ((a / b) - (st.X / st.Y)) * V[i][div] * N == A[i]
		// (a / b) * V[i][div] * N - (st.X / st.Y) * V[i][div] * N == A[i]
		// (a / b) * V[i][div] * N * st.Y - st.X * V[i][div] * N == A[i] * st.Y
		// (a / b) * V[i][div] * N * st.Y = st.X * V[i][div] * N + A[i] * st.Y
		a = 1LL * st.X * V[i][div] * N + 1LL * A[i] * st.Y;
		b = 1LL * V[i][div] * N * st.Y;
		return f(a, b);
	}
	else {
		long long sa, sb;
		sa = 1LL * lft * V[i][div]; sb = st.Y;
		tie(sa, sb) = f(sa, sb);
		for(int j=div+1; j<L; j++) {
			// sa / sb + V[i][j] >= A[i] / N
			if(1LL * sa * N + 1LL * V[i][j] * sb * N >= 1LL * A[i] * sb) {
				// (a / b) * V[i][j] + (sa / sb) == A[i] / N
				// a/b * V[i][j] * sb * N + sa * N == A[i] * sb
				// a/b * V[i][j] * sb * N == A[i] * sb - sa * N;
				b = 1LL * V[i][j] * sb * N;
				a = 1LL * b * j + 1LL * A[i] * sb - 1LL * sa * N;
				return f(a, b);
			}
			sa += 1LL * sb * V[i][j];
			tie(sa, sb) = f(sa, sb);
		}
	}
	return INF;
}

int main() {
	scanf("%d%d",&N,&L);
	for(int i=1; i<=N; i++) {
		for(int j=0; j<L; j++) {
			scanf("%d",&V[i][j]);
			A[i] += V[i][j];
		}
	}
	vector<int> S, P;
	vector<pll> ans; pll now = {0, 1};
	for(int i=1; i<=N; i++) S.push_back(i);
	for(int i=1; i<=N; i++) {
		int mni = -1; pll mn = INF;
		for(int j=0; j<S.size(); j++) {
			pll tmp = R(S[j], now);
			if(tmp < mn) mn = tmp, mni = j;
		}
		now = mn;
		ans.push_back(now);
		P.push_back(S[mni]);
		if(mni == -1) return !printf("-1");
		S.erase(S.begin() + mni);
	}
	if(now < (pll){L, 1LL} || now == (pll){L, 1LL}) {
		for(int i=0; i<N-1; i++) printf("%lld %lld\n", ans[i].X, ans[i].Y);
		for(auto& it: P) printf("%d ", it);
	}
	else puts("-1");
	return 0;
}

Compilation message

naan.cpp: In function 'int main()':
naan.cpp:74:17: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
   for(int j=0; j<S.size(); j++) {
                ~^~~~~~~~~
naan.cpp:62:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d%d",&N,&L);
  ~~~~~^~~~~~~~~~~~~~
naan.cpp:65:9: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
    scanf("%d",&V[i][j]);
    ~~~~~^~~~~~~~~~~~~~~

Subtask #1
0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 256 KB Output is correct
2 Incorrect 2 ms 376 KB Integer parameter [name=A_i] equals to -1, violates the range [1, 2000000000000]
3 Halted 0 ms 0 KB -

Subtask #2
0 / 24.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 3 ms 376 KB Output is correct
4 Incorrect 3 ms 376 KB Integer parameter [name=A_i] equals to -1, violates the range [1, 2000000000000]
5 Halted 0 ms 0 KB -

Subtask #3
0 / 71.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 256 KB Output is correct
2 Incorrect 2 ms 376 KB Integer parameter [name=A_i] equals to -1, violates the range [1, 2000000000000]
3 Halted 0 ms 0 KB -