你是岛上的穷人。这个岛上只有一家商店，除了星期天，这家商店一周的所有日子都营业。考虑以下约束：
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- N–每天可以购买的最大食物单位。
- S–生存所需的天数。
- M——每天生存所需的食物单位。
现在是星期一，你需要在接下来的几天里活下来。 找出你需要在商店购买食物的最短天数，这样你才能在接下来的几天里生存下来， 或者确定它不可能生存。 例如：
输入：S=10 N=16 M=2 输出：是2 解释1： 一个可能的解决方案是在第一天（星期一）买一盒，从这个盒子吃到第八天（星期一）就足够了。现在，在第9天（星期二），你买了另一盒巧克力，用里面的巧克力度过第9天和第10天。 输入：102030 输出：否 解释2：即使你买了食物，你也活不下去，因为你一天能买的最大单位数少于一天所需的食物。
方法： 在这个问题上，连续几天购买食物的贪婪做法是正确的方向。 如果我们能存活前7天，那么我们可以存活任意天数。为此，我们需要检查两件事 ->看看我们能否活一天。 ->（S>=7）如果我们在一周的前6天购买食物，并且我们可以在一周内生存，即我们在一周内可以购买的食物总量（6*N）大于或等于我们在一周内生存所需的食物总量（7*M），那么我们就可以生存。
注意：我们在前6天购买食物，因为我们从周一开始计算，周日商店将继续营业。 如果上述任何一种情况都不成立，那么我们就无法生存下去
该方程可推导为：
我们应该购买的食物量>=生存所需的食物量。–>方程式1
我们购买的食物数量=我们购买的次数（天）*我们一次购买的食物数量（N）
生存所需的食物量=我们生存所需的天数（S）*每天所需的食物量（M）。
现在从我们的方程式1：
天*N>=S*M
因此，天数=ceil（S*M）/N
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 <bits/stdc++.h> 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" ; 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 bartwal |
C#
// 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 Semwal |
PHP
<?php // PHP 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 // Function to find // the minimum $days function 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) && $S > 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
<script> // JavaScript 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 // function to find the minimum days function 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) && S > 6) || M > N) document.write( "No" ); else { // If we can survive then we can // buy ceil(A/N) times where A is // total units of food required. let days = (M * S) / N; if (((M * S) % N) != 0) days++; document.write( "Yes " + Math.round(days)); } } // Driver Code let S = 10, N = 16, M = 2; survival(S, N, M); // This code is contributed by splevel62. </script> |
输出：
Yes 2
时间复杂性： O（1） 空间复杂性： O（1）
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