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:给你一些秤砣,找出在不超过重质量的情况下,哪些重量是称不出来的
考虑一个特殊的情况: 给你一个1,一个2,要想称出1,很明显放一个重量为1的即可,但是天平左边放1,右边放2,同样可以称出1;
可以称出的质量还可得到这样一个式子 ; m1,m2 分别为两个秤砣重量;
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=20000;
int n;
int a[maxn];
int t[maxn];
int v[maxn];
int main()
{
while(scanf("%d",&n)!=EOF)
{
int s=0;
memset(t,0,sizeof t);
memset(v,0,sizeof v);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]),s+=a[i];
v[0]=1;
v[a[1]]=1;
for(int i=2;i<=n;++i)
{
for(int j=0;j<=s;++j)
for(int k=0;k+j<=s&&k<=a[i];k+=a[i])
{
t[abs(j-k)] += v[j];
t[j+k]+=v[j];
}
for(int j=0;j<=s;++j)
{
v[j]=t[j];
t[j]=0;
}
}
int cnt=0;
for(int i=0;i<=s;++i)
if(v[i]==0) ++cnt;
cout<<cnt<<endl;
if(cnt==0) continue;
else
{
for(int i=0;i<=s;++i)
{
if(!v[i])
{
printf("%d",i);
--cnt;
if(cnt) printf(" ");
else puts("");
}
}
}
}
}