HDOJ-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): 4327 Accepted Submission(s): 1739


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3
1 2 4
3
9 2 1

[align=left]Sample Output[/align]

0
2
4 5

[align=left]Source[/align]
HDU 2007-Spring Programming Contest

题意：
给出一些砝码，可以放在天秤的两边，问有[1,sum]中有哪些重量是不可称出来的
题解:
母函数，这里比较特殊的一点是砝码可以放在天枰的左右两端，我们可以在c2[j+k]+=c1[j]
后加多一句c2[abs(j-k)]+=c[j]...即可,假设原来的砝码都放在右端，则可以把新加的砝码放在左端，得到新重量。

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int c1[10005], c2[10005];

int main()
{
int n, sum;
int a[105];
while(~scanf("%d", &n))
{
sum = 0;
for(int i = 0; i < n; ++i)
{
scanf("%d", &a[i]);
sum += a[i];
}
memset(c1, 0, (sum+1)*sizeof(int));
memset(c2, 0, (sum+1)*sizeof(int));
c1[0] = c1[a[0]] = 1;
int end = a[0];
for(int i = 2; i <= n; ++i)
{
for(int j = 0; j <= end; ++j)
{
for(int k = 0; k <= a[i-1] && j+k <= sum; k += a[i-1])
{
c2[j+k] += c1[j];
c2[abs(j-k)] += c1[j];
}
}
end += a[i-1];
for(int j = 0; j <= end; ++j)
{
c1[j] = c2[j];
c2[j] = 0;
}
}
int cnt = 0;
for(int j = 0; j <= sum; ++j)
{
if(c1[j] == 0)
c2[cnt++] = j;
}
if(cnt == 0)
printf("0\n");
else
{
printf("%d\n", cnt);
for(int i = 0; i < cnt-1; ++i)
printf("%d ", c2[i]);
printf("%d\n", c2[cnt-1]);
}
}
return 0;
}