hdoj.1709 The Balance【母函数】 2015/03/27

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6141    Accepted Submission(s): 2506

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
#include<stdio.h>#include<string.h>int a[100010],b[100010],c[100010];int main(){int n,i,j,k,sum;while(scanf("%d",&n)!=EOF){sum = 0;memset(b,0,sizeof(b));memset(c,0,sizeof(c));for( i = 0 ; i < n ; ++i ){scanf("%d",&a[i]);sum += a[i];}b[0] = b[a[0]] = 1;for( i = 1 ; i < n ; ++i ){for( j = 0 ; j <= sum ; ++j )for( k = 0 ; j+k <= sum && k <= a[i] ; k+=a[i] ){if( j>=k ) c[j-k] += b[j];else c[k-j] += b[j];c[j+k] += b[j];}for( j = 0 ; j <= sum ; ++j ){b[j] = c[j];c[j] = 0;}}int num = 0;for( i = 1 ; i <= sum ; ++i )if( b[i] == 0 ){c[num] = i;num++;}printf("%d\n",num);for( i = 0 ; i < num ; ++i ){if( i != num-1 )printf("%d ",c[i]);else printf("%d\n",c[i]);}}return 0;}