你是一个岛上的穷人。这个岛上只有一家商店，这家商店除了星期天之外，一周中的所有日子都营业。考虑以下约束：
- N – 您每天可以购买的最大食物单位。
- S – 您需要生存的天数。
- M – 每天生存所需的食物单位。
现在是星期一，你需要在接下来的 S 天里活下来。
找出您需要从商店购买食物的最少天数，以便您可以在接下来的 S 天中生存，或者确定无法生存。
例子：
Input : S = 10 N = 16 M = 2
Output : Yes 2
Explanation 1: One possible solution is to buy a box on the first day (Monday), it’s sufficient to eat from this box up to 8th day (Monday) inclusive. Now, on the 9th day (Tuesday), you buy another box and use the chocolates in it to survive the 9th and 10th day.
Input : 10 20 30
Output : No
Explanation 2: You can’t survive even if you buy food because the maximum number of units you can buy in one day is less the required food for one day.
方法：
在这个问题上，连续早期购买食物的贪婪方法是正确的方向。
如果我们可以在前 7 天存活，那么我们可以存活任意天数，因为我们需要检查两件事
-> 看看我们能不能活过一天。
-> (S >= 7) 如果我们在一周的前 6 天购买食物并且我们可以存活一周，即我们可以在一周内购买的总食物 (6*N) 大于或等于我们的总食物需要在一周内生存（7 * M）然后我们才能生存。
注意：我们在前 6 天购买食物，因为我们从星期一开始计算，并且商店将在星期日关闭。
如果上述任何条件不成立，那么我们将无法生存，否则购买食物所需的最少天数将为 = ceil（所需食物总量/我们每天可以购买的食物单位）。
CPP
// C++ program to find the minimum days on which
// you need to buy food from the shop so that you
// can survive the next S days
#include
using namespace std;
// function to find the minimum days
void survival(int S, int N, int M)
{
// If we can not buy at least a week
// supply of food during the first week
// OR We can not buy a day supply of food
// on the first day then we can't survive.
if (((N * 6) < (M * 7) && S > 6) || M > N)
cout << "No\n";
else {
// If we can survive then we can
// buy ceil(A/N) times where A is
// total units of food required.
int days = (M * S) / N;
if (((M * S) % N) != 0)
days++;
cout << "Yes " << days << endl;
}
}
// Driver code
int main()
{
int S = 10, N = 16, M = 2;
survival(S, N, M);
return 0;
} Java
// Java program to find the minimum days on which
// you need to buy food from the shop so that you
// can survive the next S days
import java.io.*;
class GFG {
// function to find the minimum days
static void survival(int S, int N, int M)
{
// If we can not buy at least a week
// supply of food during the first
// week OR We can not buy a day supply
// of food on the first day then we
// can't survive.
if (((N * 6) < (M * 7) && S > 6) || M > N)
System.out.println("No");
else {
// If we can survive then we can
// buy ceil(A/N) times where A is
// total units of food required.
int days = (M * S) / N;
if (((M * S) % N) != 0)
days++;
System.out.println("Yes " + days);
}
}
// Driver code
public static void main(String[] args)
{
int S = 10, N = 16, M = 2;
survival(S, N, M);
}
}
// This code is contributed by vt_m.Python3
# Python3 program to find the minimum days on
# which you need to buy food from the shop so
# that you can survive the next S days
def survival(S, N, M):
# If we can not buy at least a week
# supply of food during the first week
# OR We can not buy a day supply of food
# on the first day then we can't survive.
if (((N * 6) < (M * 7) and S > 6) or M > N):
print("No")
else:
# If we can survive then we can
# buy ceil(A / N) times where A is
# total units of food required.
days = (M * S) / N
if (((M * S) % N) != 0):
days += 1
print("Yes "),
print(days)
# Driver code
S = 10; N = 16; M = 2
survival(S, N, M)
# This code is contributed by upendra bartwalC#
// C# program to find the minimum days
// on which you need to buy food from
// the shop so that you can survive
// the next S days
using System;
class GFG {
// function to find the minimum days
static void survival(int S, int N, int M)
{
// If we can not buy at least a week
// supply of food during the first
// week OR We can not buy a day
// supply of food on the first day
// then we can't survive.
if (((N * 6) < (M * 7) && S > 6) || M > N)
Console.Write("No");
else {
// If we can survive then we can
// buy ceil(A/N) times where A is
// total units of food required.
int days = (M * S) / N;
if (((M * S) % N) != 0)
days++;
Console.WriteLine("Yes " + days);
}
}
// Driver code
public static void Main()
{
int S = 10, N = 16, M = 2;
survival(S, N, M);
}
}
// This code is contributed by
// Smitha Dinesh SemwalPHP
6) || $M >$N)
echo "No";
else
{
// If we can survive then we can
// buy ceil(A/N) times where A is
// total units of food required.
$days = ($M * $S) / $N;
if ((($M * $S) % $N) != 0)
$days++;
echo "Yes " , floor($days) ;
}
}
// Driver code
$S = 10; $N = 16; $M = 2;
survival($S, $N, $M);
// This code is contributed by anuj_67
?>Javascript
输出：
Yes 2时间复杂度： O(1)
空间复杂度： O(1)