组合问题已知组合数

组合问题已知组合数_组合和问题

组合问题已知组合数

Description:

描述：

This is a standard interview problem to make some combination of the numbers whose sum equals to a given number using backtracking.

这是一个标准的面试问题，它使用回溯功能将总和等于给定数字的数字进行某种组合。

Problem statement:

问题陈述：

Given a set of positive numbers and a number, your task is to find out the combinations of the numbers from the set whose summation equals to the given number.

给定一组正数和一个数字，您的任务是从集合中找出总和等于给定数字的数字组合。

    Input:
Test case T
T no. of N values and corresponding N positive numbers and the Number.
E.g.
3
6
4 1 3 2 4 5
8
7
4 5 2 5 1 3 4
10
7
10 1 2 7 6 1 5
8
Constrains:
1 <= T <= 500
1 <= N <= 20
1 <= A[i] <= 9
1 <= Number<= 50
Output:
Print all the combination which summation equals to the given number.

Example

例

    Input:
N = 7
Set[] = 10 1 2 7 6 1 5
Number = 8
Output:
1 1 6
1 2 5
1 7
2 6

Explanation with example

举例说明

Let there is a set S of positive numbers N and a positive number.

设一组正数N和一个正数S。

Making some combinations in such a way that the summation of that combination results that given number is a problem of combination and we will solve this problem using a backtracking approach.

进行某种组合，以使该组合的总和导致给定数字是组合的问题，我们将使用回溯方法解决此问题。

    Let, 
f(i) = function to insert the i^th number into the combinational subset.

In this case, we will consider two cases to solve the problem,

在这种情况下，我们将考虑两种情况来解决该问题，

We will consider ith element into the part of our combination subset. (f(i))
我们将第ith个元素纳入我们的组合子集的一部分。 ( f(i) )
We will not consider ith element into the part of our combination subset.(not f(i))
我们不会在组合子集中考虑第ith个元素。( 不是f(i) )

And every time we will check the current sum with the number. Each of the time we will count the number of occurrence and the also the combinations.

并且每次我们将用数字检查当前总和。每次，我们将计算发生的次数以及组合。

    Let,
f(i) = function to insert the i^th number into the combinational subset.

For the input:

对于输入：

    S[] = {10, 1, 2, 7, 6, 1, 5}
Number = 8

Here in this case we will discard that edges which have a current sum greater than the given number and make a count to those numbers which are equal to the given number.

在这种情况下，我们将丢弃当前总和大于给定数字的那些边，并对等于给定数字的那些数字进行计数。

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
void combination(int* arr, int n, int num, int pos, int curr_sum, vector<vector<int> >& v, vector<int> vi, set<vector<int> >& s)
{
if (num < curr_sum)
return;
if (num == curr_sum && s.find(vi) == s.end()) {
s.insert(vi);
v.push_back(vi);
return;
}
//go for the next elements for combination
for (int i = pos; i < n; i++) {
if (curr_sum + arr[i] <= num) {
vi.push_back(arr[i]);
combination(arr, n, num, i + 1, curr_sum + arr[i], v, vi, s);
vi.pop_back();
}
}
}
//print the vector
void print(vector<vector<int> > v)
{
for (int i = 0; i < v.size(); i++) {
for (int j = 0; j < v[i].size(); j++) {
cout << v[i][j] << " ";
}
cout << endl;
}
}
void combinational_sum(int* arr, int n, int num)
{
vector<vector<int> > v;
vector<int> vi;
int pos = 0;
int curr_sum = 0;
set<vector<int> > s;
combination(arr, n, num, pos, curr_sum, v, vi, s);
print(v);
}
int main()
{
int t;
cout << "Test Case : ";
cin >> t;
while (t--) {
int n, num;
cout << "Enter the value of N : ";
cin >> n;
int arr[n];
cout << "Enter the values : ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
cout << "Enter the number : ";
cin >> num;
combinational_sum(arr, n, num);
}
return 0;
}

Output

输出量

Test Case : 3
Enter the value of N : 6
Enter the values : 4 1 3 2 4 5
Enter the number : 8
1 2 5
1 3 4
3 5
4 4
Enter the value of N : 7
Enter the values : 4 5 2 5 1 3 4
Enter the number : 10
1 2 3 4
1 4 5
2 3 5
2 4 4
5 5
Enter the value of N : 7
Enter the values : 10 1 2 7 6 1 5
Enter the number : 8
1 1 6
1 2 5
1 7
2 6