可以用数组中的所有数字组成一个可被 3 整除的数
给定一个整数数组，任务是找出是否有可能用这些数字的所有数字构造一个整数，使它能被 3 整除。如果可能，则打印“是”，否则打印“否”。 示例:
Input : arr[] = {40, 50, 90}
Output : Yes
We can construct a number which is
divisible by 3, for example 945000\.
So the answer is Yes.
Input : arr[] = {1, 4}
Output : No
The only possible numbers are 14 and 41,
but both of them are not divisible by 3,
so the answer is No.
这个想法是基于这样一个事实，即一个数可以被 3 整除，如果它的位数之和可以被 3 整除。所以我们简单地求数组元素的和。如果总和能被 3 整除，我们的答案是肯定的，否则就是否定的
卡片打印处理机（Card Print Processor 的缩写）
// C++ program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
#include <bits/stdc++.h>
using namespace std;
bool isPossibleToMakeDivisible(int arr[], int n)
{
// Find remainder of sum when divided by 3
int remainder = 0;
for (int i=0; i<n; i++)
remainder = (remainder + arr[i]) % 3;
// Return true if remainder is 0.
return (remainder == 0);
}
// Driver code
int main()
{
int arr[] = { 40, 50, 90 };
int n = sizeof(arr) / sizeof(arr[0]);
if (isPossibleToMakeDivisible(arr, n))
printf("Yes\n");
else
printf("No\n");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
import java.io.*;
import java.util.*;
class GFG
{
public static boolean isPossibleToMakeDivisible(int arr[], int n)
{
// Find remainder of sum when divided by 3
int remainder = 0;
for (int i=0; i<n; i++)
remainder = (remainder + arr[i]) % 3;
// Return true if remainder is 0.
return (remainder == 0);
}
public static void main (String[] args)
{
int arr[] = { 40, 50, 90 };
int n = 3;
if (isPossibleToMakeDivisible(arr, n))
System.out.print("Yes\n");
else
System.out.print("No\n");
}
}
// Code Contributed by Mohit Gupta_OMG <(0_o)>
Python 3
# Python program to find if it is possible
# to make a number divisible by 3 using
# all digits of given array
def isPossibleToMakeDivisible(arr, n):
# Find remainder of sum when divided by 3
remainder = 0
for i in range (0, n):
remainder = (remainder + arr[i]) % 3
# Return true if remainder is 0.
return (remainder == 0)
# main()
arr = [40, 50, 90 ];
n = 3
if (isPossibleToMakeDivisible(arr, n)):
print("Yes")
else:
print("No")
# Code Contributed by Mohit Gupta_OMG <(0_o)>
C
// C# program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
using System;
class GFG
{
public static bool isPossibleToMakeDivisible(int []arr, int n)
{
// Find remainder of sum when divided by 3
int remainder = 0;
for (int i = 0; i < n; i++)
remainder = (remainder + arr[i]) % 3;
// Return true if remainder is 0.
return (remainder == 0);
}
public static void Main ()
{
int []arr = { 40, 50, 90 };
int n = 3;
if (isPossibleToMakeDivisible(arr, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
function isPossibleToMakeDivisible($arr, $n)
{
// Find remainder of sum
// when divided by 3
$remainder = 0;
for ($i = 0; $i < $n; $i++)
$remainder = ($remainder + $arr[$i]) % 3;
// Return true if remainder is 0.
return ($remainder == 0);
}
// Driver code
$arr = array( 40, 50, 90 );
$n = sizeof($arr);
if (isPossibleToMakeDivisible($arr, $n))
echo("Yes\n");
else
echo("No\n");
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
function isPossibleToMakeDivisible(arr , n)
{
// Find remainder of sum when divided by 3
var remainder = 0;
for (i=0; i<n; i++)
remainder = (remainder + arr[i]) % 3;
// Return true if remainder is 0.
return (remainder == 0);
}
var arr = [ 40, 50, 90 ];
var n = 3;
if (isPossibleToMakeDivisible(arr, n))
document.write("Yes\n");
else
document.write("No\n");
// This code contributed by Princi Singh
</script>
输出:
Yes
时间复杂度:O(n)
版权属于：月萌API www.moonapi.com，转载请注明出处
