*HDU 1709 母函数

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7677    Accepted Submission(s): 3187

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

 

题意：

n个砝码称重，天平的两端都可以放砝码，问1~sum之间的重量中有多少是这些砝码称不出来的，sum是这些砝码的总重量。

代码：

 //a克砝码有三种状态，放在天平的左端可以认为是x^-a,不放是1，放在右边是x^a，这样(x^-a,1,x^a)。但是负的重量没法用数组存，所以
 //可以用abs(k-j)表示左右砝码的差得到c2[abs(k-j)]+=c1[j];
 #include<bits\stdc++.h>
 using namespace std;
 int n,sum,a[],c1[],c2[];
 void solve()
 {
     memset(c1,,sizeof(c1));
     memset(c2,,sizeof(c2));
     c1[]=;c1[a[]]=;   //第一个表达式的系数
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=sum;j++)
         {
             int k=;
             if(k+j<=sum)
             c2[k+j]+=c1[j];
             k=a[i];
             if(k+j<=sum)
             c2[k+j]+=c1[j];
             int tem=fabs(k-j);
             if(tem<=sum)
             c2[tem]+=c1[j];
         }
         for(int j=;j<=sum;j++)
         {
             c1[j]=c2[j];
             c2[j]=;
         }
     }
 }
 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         sum=;
         for(int i=;i<=n;i++)
         {
             scanf("%d",&a[i]);
             sum+=a[i];
         }
         solve();
         int t=,ans[];
         for(int i=;i<=sum;i++)
         if(c1[i]==)
         {
             t++;
             ans[t]=i;
         }
         printf("%d\n",t);
         if(t){
         for(int i=;i<t;i++)
         printf("%d ",ans[i]);
         printf("%d\n",ans[t]);
         }
     }
     return ;
 }