集训第五周动态规划 K题 背包

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，给你一些砝码，如3个砝码1 2 4，要求你用这些砝码称出从1到砝码之和的重量，其中不能称出的重量就要输出出来

很显然1 2 4可以称出从1到7的所有重量，输出0

9 2 1首先按左物右码可以输出1 2 3 10 11 12，而6 7 8可以是6+1=2=9，7+2=9，8+1=9

所以输出4 5

解法：可以使用DP做，dp（i）代表这个i值是否可以称出来,0代表这个可以称，1代表不可以，然后初始赋值为dp（0）=0显然重量为0是可称的，后来的判断也要先以此为基石，再把其他的所有dp（i）赋值为1，代表这个数还没开始测试，当然是不可称的状态，然后再开始具体工作

1.先按左物右码的原理开始模拟，测出那些可以由此原理测出的数字

dp（j）=min｛dp（j-a（i）），dp（j）｝

2.再按照右边放一个砝码，左边往待测物体上加砝码开始测试

dp（j）=min｛dp（j+a（i）），dp（j）｝

最后再输出那些值为1的序号

#include"iostream"
#include"cstring"
using namespace std;
int a[101];
int book[10001],ans[10001];
int m,n;

void Work()
{
     for(int i=0;i<n;i++)
     for(int j=m;j>=a[i];j--)
     {
     book[j]=min(book[j],book[j-a[i]]);
     }

     for(int i=0;i<n;i++)
     for(int j=0;j<=m-a[i];j++)
     {
     book[j]=min(book[j],book[j+a[i]]);
     }

     int top=0;
     for(int i=1;i<=m;i++)
     if(book[i]==1) ans[top++]=i;

     cout<<top<<endl;
     if(top)
     {
     for(int j=0;j<top-1;j++) cout<<ans[j]<<" ";
     cout<<ans[top-1]<<endl;
     }
}

int main()
{
    while(cin>>n)
    {
     m=0;
     memset(book,0,sizeof(book));
     for(int i=0;i<n;i++)  {cin>>a[i];m+=a[i];}
     for(int j=1;j<=m;j++) book[j]=1;
     book[0]=0;
     Work();
    }
    return 0;
}

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