以所有可能的方式执行整数分区的 C++ 程序
这是一个C++程序，用于获取给定整数的所有唯一分区，以便将分区相加得到一个整数。在这个程序中，给定一个正整数n，并生成所有可能的唯一方法来表示n为正整数之和。
算法
Begin
function displayAllUniqueParts(int m):
Declare an array to store a partition p[m].
Set Index of last element k in a partition to 0
Initialize first partition as number itself, p[k]=m
Create a while loop which first prints current partition, then generates next partition. The loop stops when the current partition has all 1s.
Display current partition as displayArray(p, k + 1);
Generate next partition:
Initialize val=0.
Find the rightmost non-one value in p[]. Also, update the val so that we know how much value can be accommodated.
If k < 0,
all the values are 1 so there are no more partitions
Decrease the p[k] found above and adjust the val.
If val is more,
then the sorted order is violeted. Divide val in different values of size p[k] and copy these values at different positions after p[k].
Copy val to next position and increment position.
End示例代码
#include输出结果using namespace std; void printArr(int p[], int m) { for (int i = 0; i < m; i++) cout << p[i] << " "; cout << endl; } void printAllUniqueParts(int m) { int p[m]; int k = 0; p[k] = m; while (true) { printArr(p, k + 1); int rem_val = 0; while (k >= 0 && p[k] == 1) { rem_val += p[k]; k--; } if (k < 0) return; p[k]--; rem_val++; while (rem_val > p[k]) { p[k + 1] = p[k]; rem_val = rem_val - p[k]; k++; } p[k + 1] = rem_val; k++; } } int main() { cout << "All Unique Partitions of 3\n"; printAllUniqueParts(3); cout << "\nAll Unique Partitions of 4\n"; printAllUniqueParts(4); cout << "\nAll Unique Partitions of 5\n"; printAllUniqueParts(5); return 0; }
All Unique Partitions of 3 3 2 1 1 1 1 All Unique Partitions of 4 4 3 1 2 2 2 1 1 1 1 1 1 All Unique Partitions of 5 5 4 1 3 2 3 1 1 2 2 1 2 1 1 1 1 1 1 1 1