hdu(1709) 母函数

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8017    Accepted Submission(s): 3336

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

题意：一个天平秤，问在他的最大范围的称量范围内，有多少种质量是无法称量的。

思路 ：砝码左右都可以放置，例如 1 9 可以称量8

输入 若无就输出0

若有就输出个数 以及 不能称量的质量

#include 
using namespace std;
int main()
{
    int n;
    int c1[10005],c2[10005];
    int a[1000],b[1000];
    while(~scanf("%d",&n))
    {
        int sum=0,ans=0;
        vector x;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        fill(c1,c1+10005,0);
        fill(c2,c2+10005,0);
        for(int i=0;i<=a[1];i+=a[1])
        c1[i]=1;
        for(int i=2;i<=n;i++)
        {
            for(int j=0;j<=sum;j++)
            {
                for(int k=0;k+j<=sum&&k<=a[i];k+=a[i])
                {
                    if(j>=k)           //左右放置的情况 
                    c2[j-k]+=c1[j];
                    else
                    c2[k-j]+=c1[j];     //左右放置的情况 
                    c2[k+j]+=c1[j];
                }
            }
            for(int j=0;j<=sum;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        for(int i=1;i<=sum;i++)
        {
            if(c1[i]==0)
            x.push_back(i);
        }
        if(x.size()==0) printf("0\n");
        else
        {
            printf("%d\n",x.size());
            for(int i=0;i