hdu 1709 The Balance

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.

 这题是母函数的解决的，可以算是变形的母函数吧，代码以及详细解答如下：

#include<stdio.h>
#include<string.h>
int c1[10001],c2[10001];
int main()
{
    int n,a[101],i,j,k,sum,count;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;count=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        c1[0]=1;c1[a[1]]=1;
        for(i=2;i<=n;i++)//表示第i个表达式，即第i个括号
        {
            for(j=0;j<=sum;j++)//j表示允许的最大值，即指数
                for(k=0;k+j<=sum&&k<=a[i];k+=a[i])//k表示被乘表达式的指数
                {
                    c2[k+j]+=c1[j];
                    if(j>k)           //与一般的母函数不同之处就在这里，从题目给出的输入案例可知，砝码可以放在左右两边
                    c2[j-k]+=c1[j];
                    else c2[k-j]+=c1[j];
                }
            for(j=0;j<=sum;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        for(i=1;i<=sum;i++)//后面都是为输出而做出的工作
             if(c1[i]==0)
                count++;
        printf("%d\n",count);
        for(i=1;i<=sum;i++)
            if(c1[i]==0&&count==1)
            {
                printf("%d\n",i);
                break;
            }
            else if(c1[i]==0) 
            {
                printf("%d ",i);
                count--;
            }
    }
    return 0;
}