HDU 1709 （母函数）

The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7565 Accepted Submission(s): 3138

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 n 种砝码，问你（1 —> s）里面哪一个重量无法测量，s 是所有砝码总和， 注意天平两端都可以放砝码
题解：用母函数的方法，这次我的母函数有点改进，第一就是 j 的查询范围变小，第二就是由于天平两端都可以放，所以跳转后距离可以是 j + k 也可以是 j - k 取绝对值，这样就把所有范围都考虑完了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 110#define M 10010#define CRL(a, b) memset(a, b, sizeof(a))int n, sum, we[N], num1[M], num2[M];int abss(int a){    return a < 0 ? (-a) : a;}void getnum(){    CRL(num1, 0);    CRL(num2, 0);    num1[0] = 1;    int flag = we[1];//定义 flag 为 j 的最大巡查范围     for(int i=1; i<=n; i++)    {        for(int j=0; j<=flag; j++)        {            for(int k=0; k+j<=sum && k<=we[i]; k+=we[i])            {                num2[k + j] += num1[j];                int t = abss(j - k);                num2[t] += num1[j];            }        }        flag += we[i];        for(int j=0; j<=flag; j++)        {            num1[j] = num2[j];            num2[j] = 0;        }    }}int main(){    while(scanf("%d", &n) != EOF)    {        sum = 0;        for(int i=1; i<=n; i++)        {            scanf("%d", &we[i]);            sum += we[i];        }        getnum();        int ans[M], flag = 0;        for(int i=1; i<=sum; i++)        {            if(!num1[i])            {                ans[flag++] = i;            }        }        if(!flag)        {            printf("0\n");        }        else        {            printf("%d\n", flag);            for(int i=0; i<flag; i++)            {                printf("%d", ans[i]);                if(i == flag-1)                {                    printf("\n");                }                else                {                    printf(" ");                }            }        }    }    return 0;}