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司法研究执行给定数字的唯一分解的 C++ 程序
这是一个用于获取给定整数的所有唯一分解的 C++ 程序，这样一组分区的加法会产生一个整数。在该程序中，给定一个正整数 n，我们将生成所有可能将 n 表示为正整数之和的唯一方法。
算法
Begin function displayAllUniqueParts(int m): 1) Set Index of last element k in a partition to 0 2) Initialize first partition as number itself, p[k]=m 3) Create a while loop which first prints current partition, then generates next partition. The loop stops when the current partition has all 1s. 4) Display current partition as displayArray(p, k + 1) 5) Generate next partition: 6) 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. 7) If val is more, then the sorted order is violated. 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
#include<iostream>
using namespace std;
void displayArray(int p[], int m) //to print the array
{
for (int i = 0; i < m; i++)
cout << p[i] << " ";
cout << endl;
}
void displayAllUniqueParts(int m)
{
int p[m];
int k = 0;
p[k] = m;
while (true)
{
displayArray(p, k + 1);
int val = 0; //initialize val
while (k >= 0 && p[k] == 1)
{
val += p[k]; //update val
k--;
}
if (k < 0)
return;
p[k]--;
val++;
while (val > p[k]) //if val is more
{
p[k + 1] = p[k];
val = val - p[k];
k++;
}
p[k + 1] = val;
k++;
}
}
int main()
{
cout << "Display All Unique Partitions of 3\n";
displayAllUniqueParts(3);
cout << "\nDisplay All Unique Partitions of 4\n";
displayAllUniqueParts(4);
cout << "\nDisplay All Unique Partitions of 5\n";
displayAllUniqueParts(5);
return 0;
}输出
Display All Unique Partitions of 3 3 2 1 1 1 1 Display All Unique Partitions of 4 4 3 1 2 2 2 1 1 1 1 1 1 Display 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
更新于：30-Jul-2019
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