查找给定电话号码的最后两位缺失数字
给定一个电话号码的八位数字为整数 N ，任务是当最后两位数字为给定的八位数字之和时，找出缺失的最后两位数字并打印出完整的号码。 例:
输入: N = 98765432 输出: 9876543244 输入: N = 10000000 输出:100000001
进场:
- 使用模 10 运算符(%10)从 N 中逐个获取电话号码的八位数字。
- 将这些数字加在一个变量中，比如 sum ，得到这八个数字的总和。
- 现在，有两种情况:
- 如果加上< 10 ，那么它就是一个位数，即在开头插入 0 ，使其成为两位数，而不影响数值。
- Else sum 是最后两位数字代表的数字。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to find the last two
// digits of the number and
// print the complete number
void findPhoneNumber(int n)
{
int temp = n;
int sum;
// Sum of the first eight
// digits of the number
while (temp != 0) {
sum += temp % 10;
temp = temp / 10;
}
// if sum < 10, then the two digits
// are '0' and the value of sum
if (sum < 10)
cout << n << "0" << sum;
// if sum > 10, then the two digits
// are the value of sum
else
cout << n << sum;
}
// Driver code
int main()
{
long int n = 98765432;
findPhoneNumber(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to find the last two
// digits of the number and
// print the complete number
static void findPhoneNumber(int n)
{
int temp = n;
int sum = 0;
// Sum of the first eight
// digits of the number
while (temp != 0)
{
sum += temp % 10;
temp = temp / 10;
}
// if sum < 10, then the two digits
// are '0' and the value of sum
if (sum < 10)
System.out.print(n + "0" + sum);
// if sum > 10, then the two digits
// are the value of sum
else
System.out.print(n +""+ sum);
}
// Driver code
public static void main(String[] args)
{
int n = 98765432;
findPhoneNumber(n);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python 3 implementation of the approach
# Function to find the last two
# digits of the number and
# print the complete number
def findPhoneNumber(n):
temp = n
sum = 0
# Sum of the first eight
# digits of the number
while (temp != 0):
sum += temp % 10
temp = temp // 10
# if sum < 10, then the two digits
# are '0' and the value of sum
if (sum < 10):
print(n,"0",sum)
# if sum > 10, then the two digits
# are the value of sum
else:
n = str(n)
sum = str(sum)
n += sum
print(n)
# Driver code
if __name__ == '__main__':
n = 98765432
findPhoneNumber(n)
# This code is contributed by Surendra_Gangwar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to find the last two
// digits of the number and
// print the complete number
static void findPhoneNumber(int n)
{
int temp = n;
int sum = 0;
// Sum of the first eight
// digits of the number
while (temp != 0)
{
sum += temp % 10;
temp = temp / 10;
}
// if sum < 10, then the two digits
// are '0' and the value of sum
if (sum < 10)
Console.Write(n + "0" + sum);
// if sum > 10, then the two digits
// are the value of sum
else
Console.Write(n + "" + sum);
}
// Driver code
static public void Main ()
{
int n = 98765432;
findPhoneNumber(n);
}
}
// This code is contributed by jit_t
java 描述语言
<script>
// Javascript implementation of the approach
// Function to find the last two
// digits of the number and
// print the complete number
function findPhoneNumber(n)
{
let temp = n;
let sum=0;
// Sum of the first eight
// digits of the number
while (temp != 0) {
sum += temp % 10;
temp = Math.floor(temp / 10);
}
// if sum < 10, then the two digits
// are '0' and the value of sum
if (sum < 10)
document.write(n + "0" + sum);
// if sum > 10, then the two digits
// are the value of sum
else
document.write(n + "" + sum);
}
// Driver code
let n = 98765432;
findPhoneNumber(n);
// This code is contributed by Mayank Tyagi
</script>
Output:
9876543244
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