The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5706 Accepted Submission(s): 2311
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
Mean:
给你N个砝码，每个砝码有自己的重量，现在要你计算从1~S中哪些重量是不能用这些砝码称出来的。(其中S为所有砝码的重量之和)。
analyse:
就是一道母函数的运用题，要注意的是砝码可以摆放在两个托盘上，所以要注意的是两个托盘两边的差值也能称出来的情况，其他的和普通母函数差不多。
Time complexity：O(n^3)
Source code：
// Memory Time
// 1347K 0MS
// by : Snarl_jsb
// 2014-09-19-11.59
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 10100
#define LL long long
using namespace std;
int val[N];
int c1[N],c2[N];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
int n;
while(cin>>n)
{
long long sum=0;
for(int i=1;i<=n;++i)
{
cin>>val[i];
sum+=val[i];
}
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
c1[0]=c1[val[1]]=1;
for(int i=2;i<=n;++i)
{
for(int j=0;j<=sum;++j)
{
for(int k=0;k<=1;++k)
{
c2[k*val[i]+j]+=c1[j];
c2[abs(k*val[i]-j)]+=c1[j];
}
}
for(int j=0;j<=sum;++j)
{
c1[j]=c2[j];
c2[j]=0;
}
}
int res[N],ans=0;
for(int i=1;i<=sum;++i)
{
if(!c1[i])
{
res[ans++]=i;
}
}
if(!ans)
{
puts("0");
continue;
}
cout<<ans<<endl;
ans--;
for(int i=0;i<ans;++i)
{
cout<<res[i]<<" ";
}
cout<<res[ans]<<endl;
}
return 0;
}
