数组部分之和问题

一个集合x有都不相同的n个元素，使用这个集合中的不定个数的元素，组成一个和为s的序列，求出所有符合的序列，元素可以重复使用，只要元素的个数相同不考虑顺序。

比如集合是x={2,3,4,5,7}; n=5, s=12可以得出以下的序列：

2       2       2       2       2       2
2       2       2       2       4
2       2       2       3       3
2       2       3       5
2       2       4       4
2       3       3       4
2       3       7
2       5       5
3       3       3       3
3       4       5
4       4       4
5       7

#include <iostream>   #include <vector>   using namespace std;  int PushVector(int *x,int n,int s,vector<int> &tablelist,int Position)  {      if(s<0)      {          tablelist.clear();          return 0;      }      else if(s==0)      {          for(int i=0;i<tablelist.size();i++)          {              cout<<tablelist[i]<<"\t";          }          cout<<"\n";                     return 0;      }      else if(s>0)      {          for(int i=Position;i<n;i++)          {              vector<int> newtablelist;              for (int j=0;j<tablelist.size();j++)              {                  newtablelist.push_back(tablelist[j]);              }              newtablelist.push_back(x[i]);              int news = s-x[i];              //cout<<i<<","<<news<<"\n";               PushVector(x,n,news,newtablelist,i);          }            }      else     {            }      return 0;  }        /* run this program using the console pauser or add your own getch, system("pause") or input loop */ int FindResult(int * x,int n,int s)  {      vector<int> tablelist;      tablelist.clear();      PushVector(x,n,s,tablelist,0);      return 0;  }              int main(int argc, char** argv)   {      int x[]= {2,3,4,5,7};      int n =5;      int s=12;      FindResult(x,n,s);      system("pause");      return 0;  }   #include <iostream>#include <vector>using namespace std;int PushVector(int *x,int n,int s,vector<int> &tablelist,int Position){    if(s<0)    {        tablelist.clear();        return 0;    }    else if(s==0)    {        for(int i=0;i<tablelist.size();i++)        {            cout<<tablelist[i]<<"\t";        }        cout<<"\n";                 return 0;    }    else if(s>0)    {        for(int i=Position;i<n;i++)        {            vector<int> newtablelist;            for (int j=0;j<tablelist.size();j++)            {                newtablelist.push_back(tablelist[j]);            }            newtablelist.push_back(x[i]);            int news = s-x[i];            //cout<<i<<","<<news<<"\n";            PushVector(x,n,news,newtablelist,i);        }      }    else    {      }    return 0;}  /* run this program using the console pauser or add your own getch, system("pause") or input loop */int FindResult(int * x,int n,int s){    vector<int> tablelist;    tablelist.clear();    PushVector(x,n,s,tablelist,0);    return 0;}    int main(int argc, char** argv) {    int x[]= {2,3,4,5,7};    int n =5;    int s=12;    FindResult(x,n,s);    system("pause");    return 0;}

#include <stdio.h>#include <stdlib.h>#include <assert.h>#include <string.h>#include <memory.h>#define _DEBUG 1#define MAXM 20#define MAXN 10int arr[MAXN];//数数组bool path[MAXN][MAXM+1];//路线bool hashtable[MAXM+1];//是否存在int c;//记录打印队列大小int v[MAXM+1];//打印队列，数组大小最大为MAXN+1，此时每个数为1void printPath(int i,int j,int c){if(path[i][j] &&j-arr[i]==0){//第一个点printf("%d ",arr[i]);for(int t=c-1;t>0;t--){printf("%d ",v[t]);}printf("%d\n",v[0]);return;}assert(path[i][j]);v[c++]=arr[i];//将元素加入到打印队列中for(int k=1;k<=i;k++){if(path[k][j-arr[i]])printPath(k,j-arr[i],c);}c--;//将元素从打印队列中删除}void solve(int size,int m){//size表示数组的大小,从1开始int i,j;//初始化memset(path,false,sizeof(path));hashtable[0]=true;for(i=1;i<=size;i++){for(j=arr[i];j<=m;j++){if(hashtable[j-arr[i]]){hashtable[j]=true;path[i][j]=true;}}}for(i=1;i<=size;i++){if(path[i][m]){printPath(i,m,0);}}}int main(){#if _DEBUG==1freopen("interview.in","r",stdin);#endifint n,m;int i,j;scanf("%d %d",&n,&m);for(i=1;i<=n;i++){scanf("%d",&arr[i]);}solve(n,m);return 0;}