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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];
}

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