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): 5750    Accepted Submission(s): 2337


[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

<span style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;">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]</span><br style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;" /><span style="font-family: 'Microsoft YaHei', 微软雅黑, Lucida, Verdana, 'Hiragino Sans GB', STHeiti, 'WenQuanYi Micro Hei', SimSun, sans-serif; font-size: 14px; line-height: 24px;">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。</span>

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
int main()
{
int c1[15000],c2[15000],a[150],count[15000];
int n,i,j,k,s,counts;
while(~scanf("%d",&n))
{
int s=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
s+=a[i];
}
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));

c1[0]=c1[a[0]]=1;
int end=a[0];
for(i=2;i<=n;i++)
{

for(j=0;j<=end;j++)
{
for(k=0;j+k<=s && k<=a[i-1];k+=a[i-1])
{
c2[j+k]+=c1[j];
c2[abs(j-k)]+=c1[j];//系数可能为负数
}
}
end+=a[i-1];
memcpy(c1,c2,sizeof(c1));
memset(c2,0,sizeof(c2));

}
int cnt=0;
for(i=0;i<=s;i++)
{
if(c1[i]==0)
{

c2[cnt++]=i;
}

}
if(cnt==0)
printf("0\n");
else
{
printf("%d\n",cnt);
for(i=0;i<cnt-1;i++)
{
printf("%d ",c2[i]);
}
printf("%d\n",c2[i]);
}
}
return 0;
}