/*The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5541 Accepted Submission(s): 2243
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
Source
HDU 2007-Spring Programming Contest
*/
#include<stdio.h>
#include<string.h>
int main()
{
int a[110], b[11000], i, j, k, n, sum, c[11000], lc;
while(scanf("%d", &n) != EOF)
{
sum = 0;
memset(b, 0 , sizeof(b));
for(i = 0; i < n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
}
b[0] = 1;//初始化
b[a[0]] = 1;//初始化
for(i = 2; i <= n; i++)
{
memset(c, 0, sizeof(c));
for(j = 0; j <= sum; j++)
for(k = 0; k + j <= sum && k <= a[i-1]; k += a[i-1])//每个重量的砝码只有一个
{
if( k > j) c[k-j] += b[j];//双向称砝码
else c[j-k] += b[j];
c[j+k] += b[j];
}
for(k =0; k <= sum ; k++)
b[k] = c[k];
}
int num = 0;
for(i = 0;i <= sum; i++)
if( !b[i])
num++;
printf("%d\n", num);
if(num == 0)
continue;
else
{
j = 0;
for(i = 0; i <= sum; i++)
if(!b[i])
{
j++;
if(j != num)
printf("%d ", i);
else
printf("%d\n", i);
}
}
}
return 0;
}
题意：给你n个砝码，用天秤秤出所能得到的各种质量，输出1到砝码总和之间所不能得到的数字。
注意：砝码可以左右分开放，得到不同的质量数。 这里用的依旧是母函数，难点在于怎么处理左右放砝码，左右放砝码就好比是减法得出一个新的数字。 | 
转播
