hdu 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): 6260 Accepted Submission(s): 2580



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

题目分析：

首先判断某种情况能不能组成，利用母函数，只不过判断时不再是单纯的相加，而是多了相减的情况，因为天平两边都可以放砝码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#define MAX 107

using namespace std;

int n;
int b[MAX*MAX];
int c[MAX];
int a[MAX*MAX];
vector<int> v,ans;

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        int sum = 0;
        memset ( c, 0 , sizeof ( c ) );
        for ( int i = 1 ; i <= n ; i++ )
        {
            scanf ( "%d" , &b[i] );
            c[b[i]]++;
            sum += b[i];
        }
        v.clear();
        for ( int i = 1 ; i <= 100 ; i++ )
            if ( c[i] ) v.push_back ( i );
        int len = v.size();
        memset ( a , 0 , sizeof ( a ) );
        for ( int i = 0 ; i <= c[v[0]] ; i++ )
            a[v[0]*i] = 1;
        for ( int i = 1 ; i < len ; i++ )
        {
            memset ( b , 0 , sizeof ( b ) );
            for ( int j = 0 ; j <= sum ; j++ )
                for ( int k = 0 ; k <= c[v[i]] ; k++ )
                    if ( a[j] )
                    {
                        if ( j+k*v[i] <= sum ) b[j+k*v[i]] = 1;
                        if ( j-k*v[i] >= 0 ) b[j-k*v[i]] = 1;
                        if ( k*v[i]-j >= 0 ) b[k*v[i]-j] = 1;
                    }
            for ( int j = 0 ; j <= sum ; j++ )
                a[j] = b[j];
        }
        ans.clear();
        for ( int i = 1 ; i <= sum ; i++ )
            if ( !a[i] ) ans.push_back ( i );
        len = ans.size();
        printf ( "%d\n" , len );
        if ( len )
        {
            printf ( "%d" , ans[0] );
            for ( int i = 1 ; i < len ; i++ )
                printf ( " %d" , ans[i] );
            puts ("");
        }
    }
}