给定正数N ，任务是检查给定数N是否可以a x + b y的形式表示 其中x和y> 1且a和b>0。如果N可以给定形式表示，则输出true，否则输出false 。
例子：
Input: N = 5
Output: true
Explanation:
5 can be expressed as 22+12
Input: N = 15
Output: false
方法：这个想法是使用完美能力的概念来确定总和是否存在。步骤如下：
- 创建一个数组(例如perfectPower [] )以存储是否为完美幂的数字。
- 现在，数组perfectPower []存储了所有构成完美幂的元素，因此我们生成了该数组中所有元素的所有可能的对和。
- 将上面步骤中计算出的总和的标记保留在数组isSum []中，因为它可以以a x + b y的形式表示 。
- 完成上述步骤后，如果isSum [N]为true，则输出true，否则输出false 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function that returns true if n
// can be written as a^m+b^n
bool isSumOfPower(int n)
{
// Taking isSum boolean array
// for check the sum exist or not
bool isSum[n + 1];
// To store perfect squares
vector perfectPowers;
perfectPowers.push_back(1);
for (int i = 0; i < (n + 1); i++) {
// Initally all sums as false
isSum[i] = false;
}
for (long long int i = 2;
i < (n + 1); i++) {
if (isSum[i] == true) {
// If sum exist then push
// that sum into perfect
// square vector
perfectPowers.push_back(i);
continue;
}
for (long long int j = i * i;
j > 0 && j < (n + 1);
j *= i) {
isSum[j] = true;
}
}
// Mark all perfect powers as false
for (int i = 0;
i < perfectPowers.size(); i++) {
isSum[perfectPowers[i]] = false;
}
// Traverse each perfectPowers
for (int i = 0;
i < perfectPowers.size(); i++) {
for (int j = i;
j < perfectPowers.size(); j++) {
// Calculating Sum with
// perfect powers array
int sum = perfectPowers[i]
+ perfectPowers[j];
if (sum < (n + 1))
isSum[sum] = true;
}
}
return isSum[n];
}
// Driver Code
int main()
{
// Given Number n
int n = 9;
// Function Call
if (isSumOfPower(n)) {
cout << "true\n";
}
else {
cout << "false\n";
}
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function that returns true if n
// can be written as a^m+b^n
static boolean isSumOfPower(int n)
{
// Taking isSum boolean array
// for check the sum exist or not
boolean []isSum = new boolean[n + 1];
// To store perfect squares
Vector perfectPowers = new Vector();
perfectPowers.add(1);
for(int i = 0; i < (n + 1); i++)
{
// Initally all sums as false
isSum[i] = false;
}
for(int i = 2; i < (n + 1); i++)
{
if (isSum[i] == true)
{
// If sum exist then push
// that sum into perfect
// square vector
perfectPowers.add(i);
continue;
}
for(int j = i * i;
j > 0 && j < (n + 1);
j *= i)
{
isSum[j] = true;
}
}
// Mark all perfect powers as false
for(int i = 0;
i < perfectPowers.size();
i++)
{
isSum[perfectPowers.get(i)] = false;
}
// Traverse each perfectPowers
for(int i = 0;
i < perfectPowers.size();
i++)
{
for(int j = i;
j < perfectPowers.size();
j++)
{
// Calculating Sum with
// perfect powers array
int sum = perfectPowers.get(i) +
perfectPowers.get(j);
if (sum < (n + 1))
isSum[sum] = true;
}
}
return isSum[n];
}
// Driver Code
public static void main(String[] args)
{
// Given number n
int n = 9;
// Function call
if (isSumOfPower(n))
{
System.out.print("true\n");
}
else
{
System.out.print("false\n");
}
}
}
// This code is contributed by amal kumar choubey Python3
# Python3 program for the above approach
# Function that returns true if n
# can be written as a^m+b^n
def isSumOfPower(n):
# Taking isSum boolean array
# for check the sum exist or not
isSum = [0] * (n + 1)
# To store perfect squares
perfectPowers = []
perfectPowers.append(1)
for i in range(n + 1):
# Initally all sums as false
isSum[i] = False
for i in range(2, n + 1):
if (isSum[i] == True):
# If sum exist then push
# that sum into perfect
# square vector
perfectPowers.append(i)
continue
j = i * i
while(j > 0 and j < (n + 1)):
isSum[j] = True
j *= i
# Mark all perfect powers as false
for i in range(len(perfectPowers)):
isSum[perfectPowers[i]] = False
# Traverse each perfectPowers
for i in range(len(perfectPowers)):
for j in range(len(perfectPowers)):
# Calculating Sum with
# perfect powers array
sum = (perfectPowers[i] +
perfectPowers[j])
if (sum < (n + 1)):
isSum[sum] = True
return isSum[n]
# Driver Code
# Given Number n
n = 9
# Function call
if (isSumOfPower(n)):
print("true")
else:
print("false")
# This code is contributed by sanjoy_62C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that returns true if n
// can be written as a^m+b^n
static bool isSumOfPower(int n)
{
// Taking isSum bool array
// for check the sum exist or not
bool []isSum = new bool[n + 1];
// To store perfect squares
List perfectPowers = new List();
perfectPowers.Add(1);
for(int i = 0; i < (n + 1); i++)
{
// Initally all sums as false
isSum[i] = false;
}
for(int i = 2; i < (n + 1); i++)
{
if (isSum[i] == true)
{
// If sum exist then push
// that sum into perfect
// square vector
perfectPowers.Add(i);
continue;
}
for(int j = i * i;
j > 0 && j < (n + 1);
j *= i)
{
isSum[j] = true;
}
}
// Mark all perfect powers as false
for(int i = 0;
i < perfectPowers.Count;
i++)
{
isSum[perfectPowers[i]] = false;
}
// Traverse each perfectPowers
for(int i = 0;
i < perfectPowers.Count;
i++)
{
for(int j = i;
j < perfectPowers.Count;
j++)
{
// Calculating Sum with
// perfect powers array
int sum = perfectPowers[i] +
perfectPowers[j];
if (sum < (n + 1))
isSum[sum] = true;
}
}
return isSum[n];
}
// Driver Code
public static void Main(String[] args)
{
// Given number n
int n = 9;
// Function call
if (isSumOfPower(n))
{
Console.Write("true\n");
}
else
{
Console.Write("false\n");
}
}
}
// This code is contributed by amal kumar choubey 输出：
true
时间复杂度： O(N)
辅助空间： O(N)