HDOJ 1709 The Balance(母函数)

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4497    Accepted Submission(s): 1808

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

 

Recommend

lcy

 

//对每个砝码可以不用，也可以放到左边或右边。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int n,m;
int a[111];
int c1[22222],c2[22222];
const int cc=10001;

int main()
{
while(scanf("%d",&n)!=EOF)
{
    int m=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        m+=a[i];
    }

    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));

    for(int i=-a[1];i<=a[1];i+=a[1])
    {
        c1[i+cc]=1;
    }

    for(int i=2;i<=n;i++)
    {
        for(int j=-m;j<=m;j++)
        {
            for(int k=-a[i];k+j<=m&&k<=a[i];k+=a[i])
            {
                c2[k+j+cc]+=c1[j+cc];
            }
        }
        for(int j=-m;j<=m;j++)
        {
            c1[j+cc]=c2[j+cc];
            c2[j+cc]=0;
        }
    }

    int aas[10000]={0};
    int ans=0;
    for(int i=0;i<=m;i++)
    {
        if(c1[i+cc]||c1[-i+cc]) ;
        else
        {
            aas[ans]=i;
            ans++;
        }
    }

    printf("%d
",ans);
    for(int i=0;i<ans;i++)
    {
        if(i==0)
            printf("%d",aas[i]);
        else
            printf(" %d",aas[i]);
    }
    if(ans)
    putchar(10);

}

    return 0;
}