看看能不能把杯子和架子排列整齐
给定三种不同类型的杯子(a[])和茶托(b[])，以及 n 个架子，看看杯子和架子是否能排列整齐。 遵循以下规则，杯碟排列整齐:
- 没有架子能同时放杯子和茶碟
- 任何架子上最多只能有 5 个杯子
- 任何架子上最多只能有 10 个调味酱
示例:
Input : a[] = {3, 2, 6}
b[] = {4, 8, 9}
n = 10
Output : Yes
Explanation :
Total cups = 11, shelves required = 3
Total saucers = 21, shelves required = 3
Total required shelves = 3 + 3 = 6,
which is less than given number of
shelves n. So, output is Yes.
Input : a[] = {4, 7, 4}
b[] = {3, 9, 10}
n = 2
Output : No
进场:整理杯子和茶托，找出杯子总数
和茶托总数
。因为同一个架子上不能超过 5 个杯子，所以通过公式
找出杯子需要的最大架子数，通过公式
找出茶碟需要的最大架子数。如果这两个值之和等于或小于
，则这种安排是可能的，否则是不可能的。
以下是上述方法的实施:
C++
// C++ code to find if neat
// arrangement of cups and
// shelves can be made
#include<bits/stdc++.h>
using namespace std;
// Function to check arrangement
void canArrange(int a[], int b[], int n)
{
int suma = 0, sumb = 0;
// Calculating total number
// of cups
for(int i = 0; i < 3; i++)
suma += a[i];
// Calculating total number
// of saucers
for(int i = 0; i < 3; i++)
sumb += b[i];
// Adding 5 and 10 so that if the
// total sum is less than 5 and
// 10 then we can get 1 as the
// answer and not 0
int na = (suma + 5 - 1) / 5;
int nb = (sumb + 10 - 1) / 10;
if(na + nb <= n)
cout << "Yes";
else
cout << "No";
}
// Driver code
int main()
{
// Number of cups of each type
int a[] = {3, 2, 6};
// Number of saucers of each type
int b[] = {4, 8, 9};
// Number of shelves
int n = 10;
// Calling function
canArrange(a, b, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to find if neat
// arrangement of cups and
// shelves can be made
import java.io.*;
class Gfg
{
// Function to check arrangement
public static void canArrange(int a[], int b[],
int n)
{
int suma = 0, sumb = 0;
// Calculating total number
// of cups
for(int i = 0; i < 3; i++)
suma += a[i];
// Calculating total number
// of saucers
for(int i = 0; i < 3; i++)
sumb += b[i];
// Adding 5 and 10 so that if
// the total sum is less than
// 5 and 10 then we can get 1
// as the answer and not 0
int na = (suma + 5 - 1) / 5;
int nb = (sumb + 10 - 1) / 10;
if(na + nb <= n)
System.out.println("Yes");
else
System.out.println("No");
}
// Driver function
public static void main(String args[])
{
// Number of cups of each type
int a[] = {3, 2, 6};
// Number of saucers of each type
int b[] = {4, 8, 9};
// Number of shelves
int n = 10;
// Calling function
canArrange(a, b, n);
}
}
Python 3
# Python code to find if neat
# arrangement of cups and
# shelves can be made
import math
# Function to check arrangement
def canArrange( a, b, n):
suma = 0
sumb = 0
# Calculating total number
# of cups
for i in range(0, len(a)):
suma += a[i]
# Calculating total number
# of saucers
for i in range(0,len(b)):
sumb += b[i]
# Adding 5 and 10 so that if
# the total sum is less than
# 5 and 10 then we can get 1
# as the answer and not 0
na = (suma + 5 - 1) / 5
nb = (sumb + 10 - 1) / 10
if(na + nb <= n):
print("Yes")
else:
print("No")
# driver function
#Number of cups of each type
a = [3, 2, 6]
# Number of saucers of each type
b = [4, 8, 9]
# Number of shelves
n = 10
#Calling function
canArrange(a ,b ,n)
# This code is contributed by Gitanjali.
C
// C# code to find if neat
// arrangement of cups and
// shelves can be made
using System;
class Gfg {
// Function to check arrangement
public static void canArrange(int []a, int []b,
int n)
{
int suma = 0, sumb = 0;
// Calculating total number
// of cups
for(int i = 0; i < 3; i++)
suma += a[i];
// Calculating total number
// of saucers
for(int i = 0; i < 3; i++)
sumb += b[i];
// Adding 5 and 10 so that if
// the total sum is less than
// 5 and 10 then we can get 1
// as the answer and not 0
int na = (suma + 5 - 1) / 5;
int nb = (sumb + 10 - 1) / 10;
if(na + nb <= n)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver function
public static void Main()
{
// Number of cups of each type
int []a = {3, 2, 6};
// Number of saucers of each type
int []b = {4, 8, 9};
// Number of shelves
int n = 10;
// Calling function
canArrange(a, b, n);
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code to find if neat
// arrangement of cups and
// shelves can be made
// Function to check arrangement
function canArrange($a, $b, $n)
{
$suma = 0; $sumb = 0;
// Calculating total number
// of cups
for( $i = 0; $i < 3; $i++)
$suma += $a[$i];
// Calculating total number
// of saucers
for( $i = 0; $i < 3; $i++)
$sumb += $b[$i];
// Adding 5 and 10 so that if the
// total sum is less than 5 and
// 10 then we can get 1 as the
// answer and not 0
$na = ($suma + 5 - 1) / 5;
$nb = ($sumb + 10 - 1) / 10;
if($na + $nb <= $n)
echo "Yes";
else
echo "No";
}
// Driver code
// Number of cups of each type
$a = array(3, 2, 6);
// Number of saucers of each type
$b = array(4, 8, 9);
// Number of shelves
$n = 10;
// Calling function
canArrange($a, $b, $n);
// This code is contributed by vt_m.
?>
java 描述语言
<script>
// Javascript code to find if neat
// arrangement of cups and
// shelves can be made
// Function to check arrangement
function canArrange( a, b, n)
{
let suma = 0, sumb = 0;
// Calculating total number
// of cups
for(let i = 0; i < 3; i++)
suma += a[i];
// Calculating total number
// of saucers
for(let i = 0; i < 3; i++)
sumb += b[i];
// Adding 5 and 10 so that if the
// total sum is less than 5 and
// 10 then we can get 1 as the
// answer and not 0
let na = Math.floor((suma + 5 - 1)/5);
let nb = Math.floor((sumb + 10 - 1)/10);
if(na + nb <= n)
document.write("Yes");
else
document.write("No");
}
// driver code
// Number of cups of each type
let a = [3, 2, 6];
// Number of saucers of each type
let b = [4, 8, 9];
// Number of shelves
let n = 10;
// Calling function
canArrange(a, b, n);
</script>
输出:
Yes
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