The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2461 Accepted Submission(s): 977
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.
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.
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 43
2 19
Sample Output
02
54
Source
又是一道关于母函数的题目。
之前没有什么思路，看了白白屋写的代码，发现还是挺好理解的
然后自己理解了之后就动手敲了下。
先开始莫名其妙的WA了，后来发现是前面初始化的时候变量没搞对
代码：
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int c1[10005],c2[10005];
int a[105],ans[10005];
int main()
{
int i,j,n,sum;
while(scanf("%d",&n)!=EOF)
{
sum=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum=sum+a[i];
}
for(i=0;i<=sum;i++)
{
c1[i]=0;
c2[i]=0;
}
c1[0]=1;
for(i=1;i<=n;i++)
{
for(j=0;j+a[i]<=sum;j++)
{
if(c1[j]==1)
{
c2[j]=1;
c2[j+a[i]]=1;
c2[abs(j-a[i])]=1;
}
}
for(j=0;j<=sum;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
int num=0;
for(i=1;i<=sum;i++)
{
if(c1[i]==0)
ans[num++]=i;
}
if(num==0)
{
printf("0\n");
continue;
}
printf("%d\n",num);
for(i=0;i<num-1;i++)
printf("%d ",ans[i]);
printf("%d\n",ans[num-1]);
}
return 0;
}