7.3.5 The Balance

The Balance

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 72 Accepted Submission(s): 50

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

意思是，用a[1],a[2]..,a[n]重量的砝码，每个砝码只能用一次，问在[1,a[1]+a[2]+..+a[n]]的范围内，有多少重量是无法通过天平称量的。

很明显，天平的话，如果是两个值 9 4 既可以称出13也可以称出5(9-4); （62ms）

只需在母函数里面加一句话，

#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

const int maxn=10210;
int cnt,n,sum,k;
int c1[maxn],c2[maxn],a[maxn],ans[maxn];

void close()
{
exit(0);
}


void init()
{
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        for (int 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 (int i=2;i<=n;i++)
        {
            for (int j=0;j<=sum;j++)
            {
                if (c1[j]==0) continue;
                k=0;
                c2[j+k]+=c1[j];
                c2[abs(j-k)]+=c1[j];
                k=a[i];
                c2[j+k]+=c1[j];
                c2[abs(j-k)]+=c1[j];
            }
            for (int j=0;j<=sum;j++)
                c1[j]=c2[j],c2[j]=0;
        }
        cnt=0;
        for (int i=1;i<=sum;i++)
            if (c1[i]==0)
            {
                cnt++;
                ans[cnt]=i;
            }
        printf("%d
",cnt);
        for (int i=1;i<=cnt;i++)
            if (i==cnt)
                printf("%d
",ans[i]);
            else
                printf("%d ",ans[i]);
    }
}

int main ()
{
    init();
    close();
    return 0;
}

另外一种方法就是DP啦，我居然还跑的最快 31ms

枚举每一个砝码，然后找出[1,sum]里面那些值已经可以取到了，然后前后更新即可

#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

const int maxn=10210;
int cnt,n,sum,k;
int a[maxn],ans[maxn],b[maxn];
bool f[maxn];

void close()
{
exit(0);
}


void init()
{
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        for (int i=1;i<=n;i++)
            scanf("%d",&a[i]),sum+=a[i];
        memset(f,false,sizeof(f));
        for (int i=1;i<=n;i++)
        {
            cnt=0;
            for (int j=sum;j>=0;j--)
                if (f[j])
                {
                    cnt++;
                    b[cnt]=j;
                }
            for (int j=1;j<=cnt;j++)
            {
                f[b[j]+a[i]]=true;
                f[abs(b[j]-a[i])]=true;
            }
            f[a[i]]=true;
        }
        cnt=0;
        for (int i=1;i<=sum;i++)
            if (not f[i])
            {
                cnt++;
                ans[cnt]=i;
            }
        printf("%d
",cnt);
        for (int i=1;i<=cnt;i++)
            if (i==cnt)
                printf("%d
",ans[i]);
            else
                printf("%d ",ans[i]);
    }
}

int main ()
{
    init();
    close();
    return 0;
}