本文主要是介绍hdu 1709 普通母函数，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目传送门：点击打开链接
这道题与别的母函数题不一样，利用了天平的原理，一开始只是盲目的套一下模版，但后来发现他们的重量不只是相加，也可以相减，但是我只是写出了一种相减的情况，就是利用的得出的重量减去第i种的重量，且没有写到第i种的重量减去得出的重量。。如果不懂可以看一下代码。
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
int n;int sum;
int c1[11000],c2[11000];
int w[110];
void init()
{sum=0;for(int i=1;i<=n;++i){cin>>w[i];sum+=w[i];}
}
void solve()
{memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));c1[w[1]]=1;c1[0]=1;for(int cnt=2;cnt<=n;++cnt){int i=w[cnt];for(int j=0;j<=sum;++j){if(c1[j]){for(int k=0;k<=i;k+=i){if(k+j<=sum)c2[k+j]+=c1[j];if(j-k>0)c2[j-k]+=c1[j];if(k>j)++c2[k-j];}}}memcpy(&c1,&c2,sizeof(c2));memset(c2,0,sizeof(c2));}vector<int> v;int num=0;for(int i=1;i<=sum;++i){if(!c1[i]){++num;v.push_back(i);}}cout<<num<<endl;if(num){for(int i=0;i<num-1;++i)cout<<v[i]<<" ";cout<<v[num-1]<<endl;}
}
int main(int argc, const char * argv[])
{while(cin>>n){init();solve();}return 0;
}
//4
//1 3 5 8
//answer:0The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7623 Accepted Submission(s): 3167
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
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