hdu1709 母函数很帅 有除得哦！！！

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
题意： 输入砝码的个数  之后输入砝码的各个质量 看从1->砝码的总和 这些质量中 那些质量测不出来 砝码可以放在天平的2边
 
 
 
 #include<stdio.h>#include<math.h>#include<string.h>int c1[10010],c2[10010],val[105],ans[10010];int main(){     int n,i,j,k,max,cnt;     while(scanf("%d",&n)!=EOF)     {         max=0;          for(i=1;i<=n;i++)          {              scanf("%d",&val[i]);              max+=val[i];          }          memset(c1,0,sizeof(c1));          memset(c2,0,sizeof(c2));         // for(i=0;i<max;i++) 注意           for(i=0;i<=val[1];i+=val[1])//是小于val[1] 因为砝码只有一个 最大为val[1]*1以前是由于砝码的数量不受限制才会小于max          {              c1[i]=1;          }          for(i=2;i<=n;i++)          {              for(j=0;j<=max;j++)                  for(k=0;k+j<=max&&k<=val[i];k=k+val[i])//k<=val[i]其实是小于 val[i]*1 因为砝码只有一个！！！                  {                      c2[k+j]+=c1[j];                      c2[abs(k-j)]+=c1[j];//这个地方也是加  因为存的是系数 也就是是方案数                  }                  for(k=0;k<=max;k++)                  {                      c1[k]=c2[k];                      c2[k]=0;                  }          }          cnt=0;          for(i=1;i<=max;i++)          {              if(!c1[i]) ans[cnt++]=i;          }          if(cnt==0) {printf("0\n");continue;}          printf("%d\n",cnt);          for(i=0;i<cnt-1;i++)              printf("%d ",ans[i]);          printf("%d\n",ans[cnt-1]);               }}