AtCoder Beginner Contest #013C – 節制

ABC013C_01.cpp

#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long int
LL n, h; 
LL a, b, c, d, e; 

// 質素な食事の最低限の回数が求められる
inline LL minCheapFood(double x){
    return ((n - x)*e - h - b*x) / (d + e) + 1;
}

int main(){
    cin >> n >> h;
    cin >> a >> b >> c >> d >> e;

    LL cost = 1e14;
    for (LL x = 0; x <= n; x++){
        LL y = max(0LL, minCheapFood(x));
        if (x + y <= n){
            cost = min(cost, (LL)(a*x + c*y));
        }
    }
    cout << cost << endl;
}

ABC013C_02.cpp

#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long int 
LL n, h; 
LL a, b, c, d, e; 

// 普通の食事の日数: x, 質素な食事の日数: y
inline bool isNotHunger(LL x, LL y){
    return h + b*x + d*y - e*(n - x - y) > 0;
}

int main(){
	cin >> n >> h;
	cin >> a >> b >> c >> d >> e;
	LL cost = 1e14;

	LL x = 0, y = 0; // 普通の食事の日数: x, 質素な食事の日数: y

	// 普通の食事だけの場合、必要となる回数を求める
	while (n > x && !isNotHunger(x, 0)) x++;

	while (1){
		//条件を満たすなら最低値を更新
		if (isNotHunger(x, y)) cost = min(cost, a*x + c*y);
		else break;

		//条件を満たさなくなるまでxを減らす(減らせないなら終了)
		if (x == 0) break;
		while (isNotHunger(x, y)) x--;

		//条件を満たすまでyを増やす(増やせないなら終了)
		if (x + y == n) break;
		while (!isNotHunger(x, y)) y++;
	}
	cout << cost << endl;
}

ABC013C_extra01.cpp

#include <iostream>
#include <algorithm>
using namespace std;

long n, h; 
int a, b, c, d, e; 

// 普通の食事の日数: x, 質素な食事の日数: y
inline bool isNotHunger(int x, int y){
    return h + b*x + d*y - e*(n - x - y) > 0;
}

int main(){
    cin >> n >> h;
    cin >> a >> b >> c >> d >> e;

    long cost = 1e10;
    for (int x = 0; x <= n; x++){
        for (int y = 0; y <= n - x; y++){
            if(isNotHunger(x, y)){
                cost = min(cost, long(a*x + c*y));
            }
        }
    }
    cout << cost << endl;
}