The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6304    Accepted Submission(s): 2592


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這個題 丫的浪費好長時間 好幾個小時 ，唉 都怪自己笨 ，用母函式來做，母函式部分算出所有可能組合出來的值得情況。假設先把最大值的組合情況放在左盤，從1開始檢測所有組合情況，這個最大值的來源有兩種情況，case1：前面某些較小的砝碼組合，case2：本身那個砝碼就最大。對於case1:那個最大值組合減去其他某種組合情況之一相當於 左盤減 右盤加！！對於 case2：相當於左盤不變，從砝碼箱裡取出其他砝碼組合加入到右盤。最坑人的是 超時好幾次！稍微優化了一下，比規定時間少2毫秒！哈哈哈

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int cnt[100002],dic[100002];
int main()
{
    int n,A[101];
    while(scanf("%d",&n)!=EOF)
    {
        int Max=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&A[i]);

            Max+=A[i];
        }

        for(int j=0; j<=Max; j++)
        {
            cnt[j]=0;
            dic[j]=0;
        }
        cnt[0]=cnt[A[0]]=1;
        for(int t=2; t<=n; t++)
        {
            for(int k=0; k<=Max; k++)
            {
                if(cnt[k])//如果前一對括號中某個指數的係數為0，那麼即使乘以下一項，也不會給當前括號內某一項指數和為T的係數有效力

                {
                    for(int h=0; h+k<=Max&&h<=A[t-1]; h+=A[t-1])
                    {
                        dic[k+h]+=cnt[k];
                    }
                }
            }
            for(int p=0; p<=Max; p++)
            {
                cnt[p]=dic[p];
                dic[p]=0;
            }

        }

        int buff[Max+1];
        memset(buff,0,sizeof(buff));
        for(int q=Max; q>=1; q--)
        {
            if(cnt[q])
            {

                for(int f=1; f<=q; f++)
                {
                    if(cnt[f])
                    {
                        if(f==q)
                            buff[q]=1;
                        else
                            buff[q-f]=1;
                    }
                }

            }
        }



        int ls[Max+1],r,count=0,t=0,flag=0;
        memset(ls,0,sizeof(ls));
        for(int a=1; a<=Max; a++)
        {
            if(!buff[a])
            {
                ls[t++]=a,count++;
                flag=1;
            }
        }
        if(!flag)
        {
            cout<<0<<endl;
            continue;
        }
        cout<<count<<endl;
        for( r=0; r<t-1; r++)
            cout<<ls[r]<<" ";
        cout<<ls[r]<<endl;
    }
    return 0;
}