hdu1790(母函数的简单应用)

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题意：给n个砝码，每个砝码的质量为Hi，[1,S]内无法用这些砝码称出的质量
这里砝码是有限的，母函数为(1 + x^h1)(1 + x^h2)(1 + x^h3).........
还有一点要注意的是砝码可以放在天平的两边，比如2,3可以称得1和5

代码如下：

#include<iostream>#include<algorithm>#include<cstring>#include<stack>#include<queue>#include<set>#include<map>#include<stdio.h>#include<stdlib.h>#include<ctype.h>#include<time.h>#include<math.h>#define inf 0x7fffffff#define eps 1e-9#define N 10005#define pi acos(-1.0)#define P system("pause")using namespace std;int c1[N],c2[N];int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);    int n;    int a[105];    while(scanf("%d",&n) != EOF)    {        int i, j,sum = 0;        for(i = 0 ; i < n;i++)        {            scanf("%d",&a[i]);            sum += a[i];        }        memset(c1,0,sizeof(c1));        memset(c2,0,sizeof(c2));        c1[0] = 1;        for(i = 0; i < n; i++)        {            for(j = 0; j <= sum; j++)            {                   c2[j] += c1[j];                   if(j+a[i] <= sum)                        c2[j+a[i]] += c1[j];                    c2[abs(j-a[i])] += c1[j];            }            for(j = 0; j <= sum; j++)            {                c1[j] = c2[j];                c2[j] = 0;            }        }        int res = 0;        for(i = 1;i <= sum; i++)            if(c1[i] == 0)                  res++;        printf("%d\n",res);        if(!res) continue;        int flag = 1;        for(i = 1; i <= sum; i++)            if(c1[i] == 0)            {                if(flag){                    printf("%d",i);                    flag = 0;                }                else printf(" %d",i);            }        printf("\n");    }    return 0;}

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