给定两个整数N和X。以这样的方式制作数字：该数字包含出现在第一个和最后一个数字
。
例子 ：
Input : N = 10, X = 5
Output : 1010101010
Explanation :
10^1 = 10
10^2 = 100
10^3 = 1000
10^4 = 10000
10^5 = 100000
Take First and Last Digit of each Power to
get required number.
Input : N = 19, X = 4
Output : 19316911
Explanation :
19^1 = 19
19^2 = 361
19^3 = 6859
19^4 = 130321
Take First and Last Digit of each Power to get required number.
方法 ：
1.逐一计算从1到X的N的所有幂。
2.将输出存储在power []数组中。
3.存储从power []数组到result []数组的power [0]，即last_digit和power [power_size – 1]，即unit_digit。
4.打印结果[]数组。
下面是上述方法的实现：
C++
// C++ program to find number formed by
// corner digits of powers.
#include
using namespace std;
// Find next power by multiplying N with
// current power
void nextPower(int N, vector &power)
{
int carry = 0;
for (int i=0 ; i < power.size(); i++)
{
int prod = (power[i] * N) + carry ;
// Store digits of Power one by one.
power[i] = prod % 10 ;
// Calculate carry.
carry = prod / 10 ;
}
while (carry)
{
// Store carry in Power array.
power.push_back(carry % 10);
carry = carry / 10 ;
}
}
// Prints number formed by corner digits of
// powers of N.
void printPowerNumber(int X, int N)
{
// Storing N raised to power 0
vector power;
power.push_back(1);
// Initializing empty result
vector res;
// One by one compute next powers and
// add their corner digits.
for (int i=1; i<=X; i++)
{
// Call Function that store power
// in Power array.
nextPower(N, power) ;
// Store unit and last digits of
// power in res.
res.push_back(power.back());
res.push_back(power.front());
}
for (int i=0 ; i < res.size(); i++)
cout << res[i] ;
}
// Driver Code
int main()
{
int N = 19 , X = 4;
printPowerNumber(X, N);
return 0 ;
} Java
// Java program to find number formed by
// corner digits of powers.
import java.io.*;
import java.util.*;
public class GFG {
static List power = new ArrayList();
// Find next power by multiplying N
// with current power
static void nextPower(Integer N)
{
Integer carry = 0;
for (int i = 0; i < power.size(); i++)
{
Integer prod = (power.get(i) * N) + carry ;
// Store digits of Power one by one.
power.set(i,prod % 10);
// Calculate carry.
carry = prod / 10 ;
}
while (carry >= 1)
{
// Store carry in Power array.
power.add(carry % 10);
carry = carry / 10 ;
}
}
// Prints number formed by corner digits of
// powers of N.
static void printPowerNumber(int X, int N)
{
// Storing N raised to power 0
power.add(1);
// Initializing empty result
List res = new ArrayList();
// One by one compute next powers and
// add their corner digits.
for (int i = 1; i <= X; i++)
{
// Call Function that store power
// in Power array.
nextPower(N) ;
// Store unit and last digits of
// power in res.
res.add(power.get(power.size() - 1));
res.add(power.get(0));
}
for (int i = 0 ; i < res.size(); i++)
System.out.print(res.get(i)) ;
}
// Driver Code
public static void main(String args[])
{
Integer N = 19 , X = 4;
printPowerNumber(X, N);
}
}
// This code is contributed by Manish Shaw
// (manishshaw1) Python3
# Python3 program to find
# number formed by
# corner digits of powers.
# Storing N raised to power 0
power = []
# Find next power by multiplying
# N with current power
def nextPower(N) :
global power
carry = 0
for i in range(0, len(power)) :
prod = (power[i] * N) + carry
# Store digits of
# Power one by one.
power[i] = prod % 10
# Calculate carry.
carry = (int)(prod / 10)
while (carry) :
# Store carry in Power array.
power.append(carry % 10)
carry = (int)(carry / 10)
# Prints number formed by corner
# digits of powers of N.
def printPowerNumber(X, N) :
global power
power.append(1)
# Initializing empty result
res = []
# One by one compute next powers
# and add their corner digits.
for i in range(1, X+1) :
# Call Function that store
# power in Power array.
nextPower(N)
# Store unit and last
# digits of power in res.
res.append(power[-1])
res.append(power[0])
for i in range(0, len(res)) :
print (res[i], end="")
# Driver Code
N = 19
X = 4
printPowerNumber(X, N)
# This code is contributed by
# Manish Shaw(manishshaw1)C#
// C# program to find number formed by
// corner digits of powers.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Collections;
class GFG {
// Find next power by multiplying N
// with current power
static void nextPower(int N, ref List power)
{
int carry = 0;
for (int i = 0; i < power.Count; i++)
{
int prod = (power[i] * N) + carry ;
// Store digits of Power one by one.
power[i] = prod % 10 ;
// Calculate carry.
carry = prod / 10 ;
}
while (carry >= 1)
{
// Store carry in Power array.
power.Add(carry % 10);
carry = carry / 10 ;
}
}
// Prints number formed by corner digits of
// powers of N.
static void printPowerNumber(int X, int N)
{
// Storing N raised to power 0
List power = new List();
power.Add(1);
// Initializing empty result
List res = new List();
// One by one compute next powers and
// add their corner digits.
for (int i = 1; i <= X; i++)
{
// Call Function that store power
// in Power array.
nextPower(N, ref power) ;
// Store unit and last digits of
// power in res.
res.Add(power.Last());
res.Add(power.First());
}
for (int i = 0 ; i < res.Count; i++)
Console.Write(res[i]) ;
}
// Driver Code
public static void Main()
{
int N = 19 , X = 4;
printPowerNumber(X, N);
}
}
// This code is contributed by Manish Shaw
// (manishshaw1) PHP
输出 ：
19316911