HDOJ 题目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): 5756    Accepted Submission(s): 2338

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

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1721 1710 1707 1792 1714 

 

思路：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。

ac代码

#include<stdio.h>#include<string.h>#include<math.h>int c1[100010],c2[100010],a[100010],b[100010];int main(){int n;while(scanf("%d",&n)!=EOF){int i,j,k,sum=0,c=0,end;for(i=0;i<n;i++){scanf("%d",&a[i]);sum+=a[i];}memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));c1[0]=c1[a[0]]=1;//end=a[0];for(i=2;i<=n;i++){for(j=0;j<=sum;j++){for(k=0;k<=a[i-1]&&k+j<=sum;k+=a[i-1]){c2[j+k] +=c1[j];c2[abs(j-k)] +=c1[j];}}//end+=a[i-1];for(j=0;j<=sum;j++){c1[j]=c2[j];c2[j]=0;}}for(i=0;i<=sum;i++){if(c1[i]==0){b[c++]=i;}}if(!c){printf("0\n");continue;}printf("%d\n",c);for(i=0;i<c;i++){if(!i)printf("%d",b[i]);elseprintf(" %d",b[i]);}printf("\n");}}