hdu 1709　The Balance（母函数）

最新推荐文章于 2019-10-06 08:10:21 发布

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#C++ #母函数

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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

Sample Output

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 ("");
        }
    }
}