Submission #100467

#	Time	Username	Problem	Language	Result	Execution time	Memory
100467		square1001	Koala Game (APIO17_koala)	C++14	37 / 100	1072 ms	512 KiB

koala

#include "koala.h"
#include <vector>
#include <algorithm>
using namespace std;

int minValue(int N, int W) {
	// TODO: Implement Subtask 1 solution here.
	// You may leave this function unmodified if you are not attempting this
	// subtask.
	int B[105], R[105];
	for(int i = 0; i < N; ++i) B[i] = (i == 0 ? 1 : 0);
	playRound(B, R);
	for(int i = 0; i < N; ++i) {
		if(R[i] <= B[i]) return i;
	}
    return -1;
}

int maxValue(int N, int W) {
	// TODO: Implement Subtask 2 solution here.
	// You may leave this function unmodified if you are not attempting this
	// subtask.
	int B[105], R[105], C[105];
	for(int i = 0; i < N; ++i) C[i] = 1;
	while(true) {
		int rem = 0;
		for(int i = 0; i < N; ++i) rem += C[i];
		if(rem == 1) break;
		for(int i = 0; i < N; ++i) {
			B[i] = (C[i] ? W / rem : 0);
		}
		playRound(B, R);
		for(int i = 0; i < N; ++i) {
			if(C[i]) C[i] = (B[i] < R[i] ? 1 : 0);
		}
	}
	for(int i = 0; i < N; ++i) {
		if(C[i]) return i;
	}
	return -1;
}
#include <iostream>
using namespace std;
int greaterValue(int N, int W) {
	// TODO: Implement Subtask 3 solution here.
	// You may leave this function unmodified if you are not attempting this
	// subtask.
	int B[105], R[105];
	vector<int> H = { 1, 2, 4, 7, 12, 22, 40, 70 };
	int T = 0;
	for(int i = 4; i >= 1; i /= 2) {
		int D = min(H[T + i] - 1, W / 2);
		for(int j = 0; j < N; ++j) {
			B[j] = (j <= 1 ? D : 0);
		}
		playRound(B, R);
		bool f0 = (B[0] < R[0]), f1 = (B[1] < R[1]);
		if(f0 != f1) return f0 ? 0 : 1;
		if(f0) T += i;
	}
    return -1;
}

int divide_val(int N, int W, int L, int R) {
	// Dividing [L, R]
	int ans = 1;
	while(true) {
		vector<pair<double, int> > v;
		for(int i = 1; i <= N; ++i) {
			if(L <= i && i <= R) {
				v.push_back(make_pair(double(i) / (ans + 1), ans + 1));
			}
			else {
				v.push_back(make_pair(i, 1));
			}
		}
		sort(v.begin(), v.end());
		reverse(v.begin(), v.end());
		int used = 0, cnt = 0;
		for(pair<double, int> i : v) {
			used += i.second;
			if(used > W) break;
			int X = i.first * (ans + 1);
			if(L <= X && X <= R) ++cnt;
		}
		if(1 <= cnt && cnt <= R - L) {
			break;
		}
		++ans;
	}
	return ans;
}

void allValues(int N, int W, int *P) {
	// TODO: Implement Subtask 4 + 5 solution here.
	// You may leave this block unmodified if you are not attempting this
	// subtask.
	int B[105], R[105];
	vector<int> rng = { 1, N + 1 };
	for(int i = 0; i < N; ++i) P[i] = 1;
	for(int i = 1; i < N; ++i) {
		int ptr = -1;
		for(int j = 0; j < i; ++j) {
			if(rng[j + 1] - rng[j] != 1) {
				ptr = j; break;
			}
		}
		int val = divide_val(N, W, rng[ptr], rng[ptr + 1] - 1);
		for(int j = 0; j < N; ++j) {
			if(P[j] == rng[ptr]) B[j] = val;
			else B[j] = 0;
		}
		playRound(B, R);
		int delta = 0;
		for(int j = 0; j < N; ++j) {
			if(P[j] == rng[ptr] && B[j] >= R[j]) ++delta;
		}
		for(int j = 0; j < N; ++j) {
			if(P[j] == rng[ptr] && B[j] < R[j]) P[j] += delta;
		}
		rng.insert(rng.begin() + ptr + 1, rng[ptr] + delta);
	}
}

• Subtask 1: 4 / 4
• Subtask 2: 15 / 15
• Subtask 3: 18 / 18
• Subtask 4: 0 / 10
• Subtask 5: 0 / 53