【HDU】1709 - The Balance（母函数）

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The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7474    Accepted Submission(s): 3101

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

 

同样是天平用母函数的问题，思路同上一篇博客。

代码如下：

#include <cstdio>#include <cstring>#define CLR(a) memset(a,0,sizeof(a))int abs(int x){return x < 0 ? -x : x;}int main(){int n;int sum;int w[111];int c[10011];int t[10011];int ant;int num[10011];while (~scanf ("%d",&n)){sum = 0;ant = 0;CLR(c);CLR(t);for (int i = 1 ; i <= n ; i++){scanf ("%d",&w[i]);sum += w[i];}c[0] = c[w[1]] = 1;for (int i = 2 ; i <= n ; i++){for (int j = 0 ; j <= sum ; j++)for (int k = 0 ; k <= w[i] && k+j <= sum ; k += w[i]){t[j+k] += c[j];t[abs(j-k)] += c[j];}for (int j = 0 ; j <= sum ; j++){c[j] = t[j];t[j] = 0;}}for (int i = 1 ; i <= sum ; i++){if (c[i] == 0)num[ant++] = i;}if (ant){printf ("%d\n",ant);for (int i = 0 ; i < ant-1 ; i++)printf ("%d ",num[i]);printf ("%d\n",num[ant-1]);}elseprintf ("0\n");}return 0;}