The Balance

The Balance

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 11

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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

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
#define inf 1<<29
int c[10010], c1[10010], c2[10010];
int main()
{
    int n,Max,Nmax;
    //freopen("F:\\1.txt", "r", stdin);
    while (scanf("%d", &n) != EOF)
    {
        int a[105]; Max = 0; Nmax = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            Max += a[i];
        }
        for (int i = 0; i < 10010; i++)  c[i] =c1[i] =c2[i] =0;    
        c2[0] = 1; c2[a[0]] = 1; 
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j <= Max; j++)
            {
                for (int k = 0; k + j <= Max&&k <= a[i]; k += a[i])
                {
                    c1[k + j] += c2[j];
                }
            }
            for (int j = 0; j <= Max; j++)
            {
                c2[j] = c1[j];
                c1[j] = 0;
            }
        }
        for (int i = 0; i <= Max; i++) c1[i] = c2[i];
        for (int i = Max; i >= 0; i--)
        {
            if (c1[i])
            {
                for (int j = i; j >= 1; j--)
                {
                    if (c1[j]) c2[i - j]=1;
                }
            }
        }            
        int cnt = 0;
        for (int i = 0; i <= Max; i++)
        {
            if (!c2[i])c[cnt++] = i;
        }
        if (cnt == 0)  printf("0\n");
        else
        {
            printf("%d\n", cnt);
            for (int i = 0; i < cnt-1; i++)
                printf("%d ", c[i]);
            printf("%d\n", c[cnt - 1]);
        }
    }
    return 0;
}