Careercup

2014-05-02 09:40

题目链接

原题：

Given a number N, write a program that returns all possible combinations of numbers that add up to N, as lists. (Exclude the N+0=N) 

For example, if N=4 return {{1,1,1,1},{1,1,2},{2,2},{1,3}}

题目：给定一个正整数N，求出所有的由正整数加起来等于N的可能组合(加数保持升序)。比如N = 4的时候，结果是{{1,1,1,1},{1,1,2},{2,2},{1,3}}。

解法：递归解决即可。

代码：

// http://www.careercup.com/question?id=6321181669982208
#include <iostream>
#include <vector>
using namespace std;

void DFS(int cur, int remain, vector<vector<int> > &result, vector<int> &sum)
{
    if (remain == 0) {
        result.push_back(sum);
        return;
    }
    if (remain < cur) {
        return;
    }
    
    int i;
    for (i = cur; i <= remain; ++i) {
        if (remain - i != 0 && remain - i < i) {
            continue;
        }
        sum.push_back(i);
        DFS(i, remain - i, result, sum);
        sum.pop_back();
    }
}

int main()
{
    int n;
    vector<int> sum;
    vector<vector<int> > result;
    int i, j;
    
    while (cin >> n && n > 0) {
        DFS(1, n, result, sum);
        
        cout << "{" << endl;
        for (i = 0; i < (int)result.size(); ++i) {
            cout << "    {";
            for (j = 0; j < (int)result[i].size(); ++j) {
                j ? cout << ", ", 1: 1;
                cout << result[i][j];
            }
            cout << "}" << endl;
        }
        cout << "}" << endl;
        
        sum.clear();
        for (i = 0; i < (int)result.size(); ++i) {
            result[i].clear();
        }
        result.clear();
    }
    
    return 0;
}