hdu1709The Balance（母函数）

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7248    Accepted Submission(s): 2990


Problem Description

Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.

 

Input

The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.

 

Output

For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.

 

Sample Input

3
1 2 4
3
9 2 1

 

Sample Output

0
2
4 5

 

Source

HDU 2007-Spring Programming Contest

个人理解题意：

给你n个砝码让你在天平上看能让这些砝码能测出小于等 于砝码的总质量，把不能测出的质量输出来。

#include <iostream>
using namespace std;
#include<stdlib.h>
#include<stdio.h>
#include <cstring>
#include<string>
#include<cmath>
#include<algorithm>
int c1[10000],c2[10000];
int main()
{
    int T;
    while(~scanf("%d",&T))
    {
        int a[10000];
        int i,sum=0,j,k,w=0;
        for(i=0;i<T;i++)
        {
                scanf("%d",&a[i]);
                sum+=a[i];//计算出砝码的的总质量
        }
        for(i=T;i<2*T;i++)
        {
            a[i]=-a[i-T];//因为砝码可以放两边
        }
        memset(c1,0,sizeof(c1));//母函数模板
         memset(c2,0,sizeof(c2));
        c1[0]=1;
        c1[a[0]]=1;
        for(i=1;i<2*T;i++)
        {
            for(k=0;k<=sum;k++)
            {
                if(c1[k]==1)//等于1说明存在。
                {
                    c2[k]=1;
                    if(k+a[i]>=0)
                    c2[k+a[i]]=1;
                }
            }
            for(j=0;j<=sum;j++)
            {
                 c1[j]=c2[j];
                 c2[j]=0;
            }
        }
        int q=0;
        for(i=1;i<=sum;i++)
        {
            if(c1[i]==0)
            {
                q++;
            }
        }
        if(q==0)
            printf("0\n");
        else
        {
            printf("%d\n",q);
            for(i=1;i<=sum;i++)
            {
                if(c1[i]==0&&w==0)//w控制格式的
               {
                printf("%d",i);
                w=1;
                }
              else  if(c1[i]==0&&w==1)
               {
                printf(" %d",i);

                }
            }
            printf("\n");
        }
    }

    return 0;
}