回溯法应用：1，2，5，10四个数任意次数相加得到一个数N

#include <iostream>  
    #include <vector>  
      
    using namespace std;  
    vector<int> vec;  
    const int a[4] = {1, 2, 5, 10};  
      
    
      
    void backup(int N)
    {  
        if (N == 0)  
        {  
            vector<int>::iterator it = vec.begin();  
            for (; it != vec.end(); ++it)  
                cout<< *it <<" ";  
            cout<<endl;  
            return;  
        }  
      
        if (N < 0) return;  
      
        for (int i = 0; i < 4; ++i)  
        {  
            if (vec.empty() || a[i] >= vec.back())// 非降序,为了去掉重复的组合  
            {  
                vec.push_back(a[i]);  
                backup(N-a[i]);  
                vec.pop_back();  
            }  
        }  
    }  
      
    int main(int argc, char* argv[])  
    {  
        backup(20);  
        return 0;  
    }

#include <iostream>
#include <string>   
using namespace std;

int count = 0;

void printAll(int* a, int n, int diff, int* path, int cur)
{
    if(diff == 0)
    {
        count++;
        for(int i = 0; i < cur; i++)
        {
            cout << path[i] << " ";
        }
        cout << endl;
        return;
    }
    for(int i = 0; i < n; i++)
    {
        if( (cur == 0 || a[i] >= path[cur-1]) && a[i] <= diff )
        {
            path[cur] = a[i];
            printAll(a, n, diff-a[i], path, cur+1);
        }
    }
}

int main()  
{  
    int a[] = {1,2,5,10};
    int b = 20;
    int* path = new int[30];
    printAll(a, 4, b, path, 0);
    cout << count << endl;
    return 0;  
}