HDOJ1709 The Balance(母函数)

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4327    Accepted Submission(s): 1739

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

 

题意：
给出一些砝码，可以放在天秤的两边，问有[1,sum]中有哪些重量是不可称出来的
题解:
母函数，这里比较特殊的一点是砝码可以放在天枰的左右两端，我们可以在c2[j+k]+=c1[j]
后加多一句c2[abs(j-k)]+=c[j]...即可,假设原来的砝码都放在右端，则可以把新加的砝码放在左端，得到新重量。

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int c1[10005], c2[10005];

int main()
{
    int n, sum;
    int a[105];
    while(~scanf("%d", &n))
    {
        sum = 0;
        for(int i = 0; i < n; ++i)
        {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        memset(c1, 0, (sum+1)*sizeof(int));
        memset(c2, 0, (sum+1)*sizeof(int));
        c1[0] = c1[a[0]] = 1;
        int end = a[0];
        for(int i = 2; i <= n; ++i)
        {
            for(int j = 0; j <= end; ++j)
            {
                for(int k = 0; k <= a[i-1] && j+k <= sum; k += a[i-1])
                {
                    c2[j+k] += c1[j];
                    c2[abs(j-k)] += c1[j];
                }
            }
            end += a[i-1];
            for(int j = 0; j <= end; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        int cnt = 0;
        for(int j = 0; j <= sum; ++j)
        {
            if(c1[j] == 0)
                c2[cnt++] = j;
        }
        if(cnt == 0)
            printf("0\n");
        else
        {
            printf("%d\n", cnt);
            for(int i = 0; i < cnt-1; ++i)
                printf("%d ", c2[i]);
            printf("%d\n", c2[cnt-1]);
        }
    }
    return 0;
}