hdu 1709 母函数

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7220    Accepted Submission(s): 2980


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：有一个天平来称物体重量，有n种砝码，每种砝码只有一个，由于每个砝码都可以放在左右两边，所以不仅要相加还要相减。

用到母函数（1+x^n1）*(1+x^n2)*……*(1+x^nn),由于能力有限只能计算出那个物体能称出来，不能算出方法有几种。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int a[10001];
int c1[10001],c2[10001];
int c3[10001];
int main()
{
int n;
while(scanf("%d",&n)==1)
{
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
int sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
c1[0]=1;c1[a[1]]=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])
{
c2[abs(j-k)]+=c1[j];
c2[j+k]+=c1[j];
}
}
for(int j=0;j<=sum;j++)
{
c1[j]=c2[j];c2[j]=0;
}
}
int cnt=0;
for(int i=1;i<=sum;i++)
{
if(c1[i]==0)
{
c3[++cnt]=i;
}
}
if(cnt==0)
{
printf("0\n");
}
else{
printf("%d\n",cnt);
for(int i=1;i<=cnt;i++)
{
if(i!=1)
printf(" ");
printf("%d",c3[i]);
}
printf("\n");
}
}
return 0;
}