Addition Decomposition Scheme

The problem is described as a positive integer N, and the decomposition method of all its positive integer addition is output.

Note:

1. Determine whether the location of the switching addition is considered as a different Decomposition Scheme Based on the input requirements.

2. Non-decomposition is also considered as a decomposition solution.

3. Output all decomposition schemes in Lexicographic Order. For the format, see the example.
The input format contains one row, which contains two positive integers, N and M, separated by a space. N indicates the positive integer to be decomposed. m is 1 or 2:

1 indicates whether the location of the switching addition is considered as a different Decomposition Scheme;

2 indicates whether the location of the switching addition is regarded as the same Decomposition Scheme.
The output format outputs several rows. Each row represents a Decomposition Scheme. For a scheme, n is output first, and then "=" is output ". Then, the number of splits is output. Different numbers are connected by a "+.
Example input 5 2
Sample output 5 = 1 + 1 + 1 + 1 + 1

5 = 1 + 1 + 1 + 2

5 = 1 + 1 + 3

5 = 1 + 2 + 2

5 = 1 + 4

5 = 2 + 3

5 = 5
Input and Output Example 2 example input 5 1
Sample output 5 = 1 + 1 + 1 + 1 + 1

5 = 1 + 1 + 1 + 2

5 = 1 + 1 + 2 + 1

5 = 1 + 1 + 3

5 = 1 + 2 + 1 + 1

5 = 1 + 2 + 2

5 = 1 + 3 + 1

5 = 1 + 4

5 = 2 + 1 + 1 + 1

5 = 2 + 1 + 2

5 = 2 + 2 + 1

5 = 2 + 3

5 = 3 + 1 + 1

5 = 3 + 2

5 = 4 + 1

5 = 5

Answer: BFS

# Include <iostream> using namespace STD; int sum = 0;/** M = 1. When the position of the switching addition is regarded as different decomposition schemes, count = 0 */void dfs1 (int n, int count, int * P) {If (n = 0) {// recursive base, result cout <sum <"="; for (INT I = 0; I <count; I ++) if (I = count-1) cout <p [I] <Endl; elsecout <p [I] <"+"; return ;}for (INT I = 1; I <= N; I ++) {P [count] = I; dfs1 (n-I, Count + 1, P);} return ;}/ * m = 2, when the position of the switching addition is regarded as the same Decomposition Scheme, Count = 0 */void dfs2 (int n, int count, int * P) {If (n = 0) {// recursive base, output to the leaf node Result cout <sum <"="; for (INT I = 0; I <count; I ++) if (I = count-1) cout <p [I] <Endl; elsecout <p [I] <"+"; return;} int K = count> 0? P [count-1]: 1; // The number in the array should be in a non-descending order for (INT I = K; I <= N; I ++) {P [count] = I; dfs2 (n-I, Count + 1, p) ;}} int main () {int n, m; cin> N> m; sum = N; int * P = new int [N]; If (M = 1) dfs1 (n, 0, P ); else if (M = 2) dfs2 (n, 0, P); Return 0 ;}