检查一个大数是否能被 7 整除
给你一个 n 位数的大数字，你要检查它是否能被 7 整除。 数字形式为(ar-1 ar-2…a2 a1 A0)的一个(r+1)位整数 n 可被 7 整除当且仅当交替数列(a2 a1 A0)–(a5 a4 a3)+(A8 a7 a6)–……可被 7 整除。
括号内的三位数代表数字形式的 3 位数。
给定的数 n 可以写成 1000 次幂的和，如下所示。 n =(a2 a1 A0)+(a5 a4 a3) 1000+(A8 a7 a6)(1000 * 1000)+…。 As 1000 = (-1)(mod 7)，根据同余关系为 1000。 对于一个正整数 n，如果两个数 a 和 b 的差 (a–b)是 n 的整数倍(也就是说，如果有一个整数 k 使得 a–b = kn)，则称这两个数 a 和 b 是全等模 n。当 a 和 b 是整数时，通常考虑这种同余关系，表示为
因此我们可以写: n = {(a2a1a 0)+(a5a4a 3)(-1)+(a8a 7a 6)(-1)*(-1)+…..}(mod 7)， 因此，当且仅当数列可被 7 整除时，n 可被 7 整除。
因此我们可以写:
n = {(a2a1a 0)+(a5a4a 3)(-1)+(a8a 7a 6)(-1)*(-1)+…..}(mod 7)，
因此，当且仅当数列可被 7 整除时，n 可被 7 整除。示例:
Input : 8955795758
Output : Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(008)(955)(795)(758)
Now, 758- 795 + 955 - 8 = 910, which is
divisible by 7
Input : 100000000000
Output : Not Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(100)(000)(000)(000)
Now, 000- 000 + 000 - 100 = -100, which is
not divisible by 7
请注意，n 中的位数不能是 3 的倍数。在这种情况下，我们在取出所有三元组(从 n 的右侧)后，在剩余数字的左侧传递零(s)以形成最后一个三元组。 一种简单高效的方法是以字符串的形式进行输入(如果需要的话，通过在数字的左边加上 0，使其长度为 3*m 的形式)，然后你必须从右到左以三为单位将数字相加，直到它变成一个 3 位数，以形成一个交替序列，并检查该序列是否可被 7 整除。
这里检查 7 的可除性的程序实现完成了。
C++
// C++ code to check divisibility of a
// given large number by 7
#include<bits/stdc++.h>
using namespace std;
int isdivisible7(string num)
{
int n = num.length(), gSum=0;
if (n == 0)
return 1;
// Append required 0s at the beginning.
if (n % 3 == 1) {
num="00" + num;
n += 2;
}
else if (n % 3 == 2) {
num= "0" + num;
n++;
}
// add digits in group of three in gSum
int i, GSum = 0, p = 1;
for (i = n - 1; i >= 0; i--) {
// group saves 3-digit group
int group = 0;
group += num[i--] - '0';
group += (num[i--] - '0') * 10;
group += (num[i] - '0') * 100;
gSum = gSum + group * p;
// generate alternate series of plus
// and minus
p *= (-1);
}
return (gSum % 7 == 0);
}
// Driver code
int main()
{
// Driver method
string num= "8955795758";
if (isdivisible7(num))
cout << "Divisible by 7";
else
cout << "Not Divisible by 7";
return 0;
}
// This code is contributed
// by Akanksha Rai
C
// C code to check divisibility of a
// given large number by 7
#include <stdio.h>
#include <string.h>
int isdivisible7(char num[])
{
int n = strlen(num), gSum=0;
char final[n+3];
if (n == 0 && num[0] == '\n')
return 1;
// Append required 0s at the beginning.
if (n % 3 == 1) {
final[0]='0';
final[1]='0';
strcat(final,num);
n += 2;
}
else if (n % 3 == 2) {
final[0]='0';
strcat(final,num);
n++;
}
// add digits in group of three in gSum
int i, GSum = 0, p = 1;
for (i = n - 1; i >= 0; i--) {
// group saves 3-digit group
int group = 0;
group += final[i--] - '0';
group += (final[i--] - '0') * 10;
group += (final[i] - '0') * 100;
gSum = gSum + group * p;
// generate alternate series of plus
// and minus
p *= (-1);
}
return (gSum % 7 == 0);
}
// Driver code
int main()
{
// Driver method
char num[] = "8955795758";
if (isdivisible7(num))
printf("Divisible by 7");
else
printf("Not Divisible by 7");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to check divisibility of a given large number by 7
class Test {
// Method to check divisibility
static boolean isDivisible7(String num)
{
int n = num.length();
if (n == 0 && num.charAt(0) == '0')
return true;
// Append required 0s at the beginning.
if (n % 3 == 1)
num = "00" + num;
if (n % 3 == 2)
num = "0" + num;
n = num.length();
// add digits in group of three in gSum
int gSum = 0, p = 1;
for (int i = n - 1; i >= 0; i--) {
// group saves 3-digit group
int group = 0;
group += num.charAt(i--) - '0';
group += (num.charAt(i--) - '0') * 10;
group += (num.charAt(i) - '0') * 100;
gSum = gSum + group * p;
// generate alternate series of plus and minus
p = p * -1;
}
// calculate result till 3 digit sum
return (gSum % 7 == 0);
}
// Driver method
public static void main(String args[])
{
String num = "8955795758";
System.out.println(isDivisible7(num) ? "Divisible by 7" : "Not Divisible by 7");
}
}
Python 3
# Python 3 code to check divisibility
# of a given large number by 7
def isdivisible7(num):
n = len(num)
if (n == 0 and num[0] == '\n'):
return 1
# Append required 0s at the beginning.
if (n % 3 == 1) :
num = "00" + str(num)
n += 2
elif (n % 3 == 2) :
num = "0" + str(num)
n += 1
# add digits in group of three in gSum
GSum = 0
p = 1
i = n-1
while i>=0 :
# group saves 3-digit group
group = 0
group += ord(num[i]) - ord('0')
i -= 1
group += (ord(num[i]) - ord('0')) * 10
i -= 1
group += (ord(num[i]) - ord('0')) * 100
GSum = GSum + group * p
# generate alternate series of
# plus and minus
p *= (-1)
i -= 1
return (GSum % 7 == 0)
# Driver code
if __name__ == "__main__":
num = "8955795758"
if (isdivisible7(num)):
print("Divisible by 7")
else :
print("Not Divisible by 7")
# This code is contributed by ChitraNayal
C
// C# code to check divisibility of a
// given large number by 7
using System;
class GFG {
// Method to check divisibility
static bool isDivisible7(String num)
{
int n = num.Length;
if (n == 0 && num[0] == '0')
return true;
// Append required 0s at the beginning.
if (n % 3 == 1)
num = "00" + num;
if (n % 3 == 2)
num = "0" + num;
n = num.Length;
// add digits in group of three in gSum
int gSum = 0, p = 1;
for (int i = n - 1; i >= 0; i--) {
// group saves 3-digit group
int group = 0;
group += num[i--] - '0';
group += (num[i--] - '0') * 10;
group += (num[i] - '0') * 100;
gSum = gSum + group * p;
// generate alternate series
// of plus and minus
p = p * -1;
}
// calculate result till 3 digit sum
return (gSum % 7 == 0);
}
// Driver code
static public void Main()
{
String num = "8955795758";
// Function calling
Console.WriteLine(isDivisible7(num) ? "Divisible by 7" : "Not Divisible by 7");
}
}
// This code is contributed by Ajit.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code to check divisibility of
// a given large number by 7
// Function to check divisibility
function isDivisible7($num)
{
$n = strlen($num) ;
if ($n == 0 && $num[0] == '0')
return true;
// Append required 0s at the beginning.
if ($n % 3 == 1)
$num = "00" . $num;
if ($n % 3 == 2)
$num = "0". $num;
$n = strlen($num);
// add digits in group of three in gSum
$gSum = 0 ;
$p = 1;
for ($i = $n - 1; $i >= 0; $i--)
{
// group saves 3-digit group
$group = 0;
$group += $num[$i--] - '0';
$group += ($num[$i--] - '0') * 10;
$group += ($num[$i] - '0') * 100;
$gSum = $gSum + $group * $p;
// generate alternate series
// of plus and minus
$p = $p * -1;
}
// calculate result till 3 digit sum
return ($gSum % 7 == 0);
}
// Driver Code
$num = "8955795758";
echo (isDivisible7($num) ?
"Divisible by 7" :
"Not Divisible by 7");
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript code to check divisibility of
// a given large number by 7
// Function to check divisibility
function isDivisible7(num)
{
let n = num.length;
if (n == 0 && num[0] == '0')
return true;
// Append required 0s at the beginning.
if (n % 3 == 1)
num = "00" + num;
if (n % 3 == 2)
num = "0" + num;
n = num.length;
// Add digits in group of three in gSum
gSum = 0 ;
let p = 1;
for(let i = n - 1; i >= 0; i--)
{
// Group saves 3-digit group
group = 0;
group += num[i--] - '0';
group += (num[i--] - '0') * 10;
group += (num[i] - '0') * 100;
gSum = gSum + group * p;
// Generate alternate series
// of plus and minus
p = p * -1;
}
// Calculate result till 3 digit sum
return (gSum % 7 == 0);
}
// Driver Code
let num = "8955795758";
document.write(isDivisible7(num) ?
"Divisible by 7" :
"Not Divisible by 7");
// This code is contributed by _saurabh_jaiswal
</script>
输出:
Divisible by 7
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