HDU 1709 The Balance（母函数）

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8192    Accepted Submission(s): 3430

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

31 2 439 2 1

 

Sample Output

024 5

 

Source

HDU 2007-Spring Programming Contest 

 

题意：

给你几个砝码，一个天平，求出不能称出的重量，砝码可以左右放。

POINT：

母函数，其实感觉就是个DP而已。天平可以放两边就是减一下。

#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>#include <algorithm>#include <map>using namespace std;#define LL long longconst int N = 12000+6;int c1[N],c2[N];int a[N];int main(){    int n;    while(~scanf("%d",&n))    {        int sum=0;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]),sum+=a[i];        memset(c1,0,sizeof c1);        c1[0]=c1[a[1]]=1;        for(int i=2;i<=n;i++)        {            for(int j=sum;j>=0;j--)            {                if(c1[j]==1)                {                    c2[j]=1;                    c2[j+a[i]]=1;                    c2[abs(j-a[i])]=1;                }            }            for(int j=0;j<=sum;j++) c1[j]=c2[j],c2[j]=0;        }        int cnt=0;        int ans[N];        for(int i=1;i<=sum;i++)        {            if(c1[i]==0)            {                ans[++cnt]=i;            }        }        if(cnt==0) printf("0\n");        else        {            printf("%d\n",cnt);            for(int i=1;i<=cnt;i++)            {                if(!(i-1)) printf("%d",ans[i]);                else printf(" %d",ans[i]);            }            printf("\n");        }    }    return 0;}