HDU 1709 母函数天平问题 可出现减法的情况 The Balance

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6652    Accepted Submission(s): 2730

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

 

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#include<stdio.h>
#include<string.h>
#include<math.h>
int num[10005];
int ans[10005];
int c[10005],temp[10005];
int main(){
   int n;
   while(scanf("%d",&n)!=EOF){
      memset(num,0,sizeof(num));
      memset(c,0,sizeof(c));
      memset(ans,0,sizeof(ans));
      memset(temp,0,sizeof(temp));
      int total=0;
      for(int i=0;i<n;i++){
        scanf("%d",&num[i]);
          total+=num[i];
        }
      for(int i=0;i<=num[0];i+=num[0])
        c[i]=1;
      for(int i=1;i<n;i++){
        for(int j=0;j<=total;j++){
            for(int k=0;k+j<=total&&k/num[i]<=1;k+=num[i]){///注意《=1
                temp[j+k]+=c[j];
                temp[(int)fabs(j-k)]+=c[j];//注意此刻应当考虑负值的情况
            }

        }
        for(int ii=0;ii<=total;ii++)
            c[ii]=temp[ii];

      }

    int cnt=0;
    for(int i=1;i<=total;i++){
        if(!c[i]){
            cnt++;
            ans[cnt]=i;
        }
    }
    printf("%d
",cnt);
    for(int i=1;i<=cnt;i++){
        printf("%d%c",ans[i],i==cnt?'
':' ');
    }
   }
   return 0;
}