子集合数（定长解）

用定长的方法求解：

 

#include<iostream>
using namespace std;

long Judge(int k,int *X,int *array)
{
    int i;              //判断计算函数
       long s=0;
       for(i=1;i<=k;i++)
           if(X[i]!=0)
               s+=array[i];
           return s;             
}


void Solution(int n,int *array,int result)
{                                    //子集和数问题解向量长度固定的结构  
    int i,j;
    int *X=new int [n+1];
    long sum;
    i=1;
       X[i]=1;
       while(i>=1)
       {      
           sum=Judge(i,X,array);
           
           if(sum<result)   //当搜索的解的和小于所给定的和数
           { 
               if(i==n)
                   X[i]=0;
               
               if(i<n)
               {
                   i++;
                   X[i]=1;
               }
           }
           
           if(sum==result)   //当搜索的解的和等于所给定的和数
           { 
               if(i<n)
               {
                   i++;
                   X[i]=0;    
               } 
               
               if(i==n)
               {  
                   for(j=1;j<=n;j++)
                       cout<<X[j]<<" ";
                   cout<<endl;
                   X[i]=0;    
               }
               
           }
           
           if(sum>result) //当搜索的解的和大于所给定的和数
           {      
               X[i]=0; 
               if(i<n)
               {
                   i++;
                   X[i]=1;
               }
           }     
           
           
           if(i==n&&X[i]==0)   //回溯
           {
               while(X[i]==0&&i>=1)
                   i--;
               X[i]=0;
               if(i>=1)
               {
                   i++;
                   X[i]=1;
               }
           }               
       }                                  
}

void main()
{
    int n,i,result;
    cin>>n>>result;
    int *array=new int [n+1];
    for(i=1;i<=n;i++)
        cin>>array[i];
    Solution(n,array,result);
}

 

测试数据：

5 15

3 5 7 10 12

1 1 1 0 0

1 0 0 0 1

0 1 0 1 0