The Balance（母函数）

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5655    Accepted Submission(s): 2285

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

31 2 439 2 1

 

Sample Output

024 5

 

Source

HDU 2007-Spring Programming Contest

模板题：

AC代码：

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;const int M = 10005;int a[110],b[M],c[M];int main(){    int n,sum,ans,res,to;    while(~scanf("%d",&n))    {        sum = 0;        for(int i = 0; i < n; i++)        {            scanf("%d",a + i);            sum += a[i];        }        memset(b,0,(sum + 1) * sizeof(int));        memset(c,0,(sum + 1) * sizeof(int));        b[0] = b[a[0]] = 1;        ans = a[0];        res = to = 0;        for(int i = 1; i < n; i++)        {            for(int j = 0; j <= ans; j++)            {                for(int k = 0; k + j <= sum && k <= a[i]; k += a[i])                {                    int t = abs(j - k);                    c[j + k] += b[j];                    c[t] += b[j];                }            }            ans += a[i];            for(int j = 0; j <= ans; j++)            {                b[j] = c[j];                c[j] = 0;            }        }        for(int i = 1; i <= sum; i++)        {            if(b[i] == 0)            {                a[to++] = i;                res++;            }        }        printf("%d\n",res);        for(int i = 0; i < to; i++)            printf(i == to - 1 ? "%d\n" : "%d ",a[i]);    }    return 0;}