# 2787. Ways to Express an Integer as Sum of Powers

## Description

Given two positive integers n and x.

Return the number of ways n can be expressed as the sum of the xth power of unique positive integers, in other words, the number of sets of unique integers [n1, n2, ..., nk] where n = n1x + n2x + ... + nkx.

Since the result can be very large, return it modulo 109 + 7.

For example, if n = 160 and x = 3, one way to express n is n = 23 + 33 + 53.

Example 1:

Input: n = 10, x = 2
Output: 1
Explanation: We can express n as the following: n = 32 + 12 = 10.
It can be shown that it is the only way to express 10 as the sum of the 2nd power of unique integers.


Example 2:

Input: n = 4, x = 1
Output: 2
Explanation: We can express n in the following ways:
- n = 41 = 4.
- n = 31 + 11 = 4.


Constraints:

• 1 <= n <= 300
• 1 <= x <= 5

## Solutions

• class Solution {
public int numberOfWays(int n, int x) {
final int mod = (int) 1e9 + 7;
int[][] f = new int[n + 1][n + 1];
f[0][0] = 1;
for (int i = 1; i <= n; ++i) {
long k = (long) Math.pow(i, x);
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (k <= j) {
f[i][j] = (f[i][j] + f[i - 1][j - (int) k]) % mod;
}
}
}
return f[n][n];
}
}

• class Solution {
public:
int numberOfWays(int n, int x) {
const int mod = 1e9 + 7;
int f[n + 1][n + 1];
memset(f, 0, sizeof(f));
f[0][0] = 1;
for (int i = 1; i <= n; ++i) {
long long k = (long long) pow(i, x);
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (k <= j) {
f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
}
}
}
return f[n][n];
}
};

• class Solution:
def numberOfWays(self, n: int, x: int) -> int:
mod = 10**9 + 7
f = [[0] * (n + 1) for _ in range(n + 1)]
f[0][0] = 1
for i in range(1, n + 1):
k = pow(i, x)
for j in range(n + 1):
f[i][j] = f[i - 1][j]
if k <= j:
f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod
return f[n][n]


• func numberOfWays(n int, x int) int {
const mod int = 1e9 + 7
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, n+1)
}
f[0][0] = 1
for i := 1; i <= n; i++ {
k := int(math.Pow(float64(i), float64(x)))
for j := 0; j <= n; j++ {
f[i][j] = f[i-1][j]
if k <= j {
f[i][j] = (f[i][j] + f[i-1][j-k]) % mod
}
}
}
return f[n][n]
}

• function numberOfWays(n: number, x: number): number {
const mod = 10 ** 9 + 7;
const f: number[][] = Array(n + 1)
.fill(0)
.map(() => Array(n + 1).fill(0));
f[0][0] = 1;
for (let i = 1; i <= n; ++i) {
const k = Math.pow(i, x);
for (let j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (k <= j) {
f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
}
}
}
return f[n][n];
}