The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7677 Accepted Submission(s): 3187
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
3
1 2 4
3
9 2 1
Sample Output
0
2
4 5
Source
题意：
n个砝码称重，天平的两端都可以放砝码，问1~sum之间的重量中有多少是这些砝码称不出来的，sum是这些砝码的总重量。
代码：
1 //a克砝码有三种状态，放在天平的左端可以认为是x^-a,不放是1，放在右边是x^a，这样(x^-a,1,x^a)。但是负的重量没法用数组存，所以 2 //可以用abs(k-j)表示左右砝码的差得到c2[abs(k-j)]+=c1[j]; 3 #include<bitsstdc++.h> 4 using namespace std; 5 int n,sum,a[102],c1[10004],c2[10004]; 6 void solve() 7 { 8 memset(c1,0,sizeof(c1)); 9 memset(c2,0,sizeof(c2)); 10 c1[0]=1;c1[a[1]]=1; //第一个表达式的系数 11 for(int i=2;i<=n;i++) 12 { 13 for(int j=0;j<=sum;j++) 14 { 15 int k=0; 16 if(k+j<=sum) 17 c2[k+j]+=c1[j]; 18 k=a[i]; 19 if(k+j<=sum) 20 c2[k+j]+=c1[j]; 21 int tem=fabs(k-j); 22 if(tem<=sum) 23 c2[tem]+=c1[j]; 24 } 25 for(int j=0;j<=sum;j++) 26 { 27 c1[j]=c2[j]; 28 c2[j]=0; 29 } 30 } 31 } 32 int main() 33 { 34 while(scanf("%d",&n)!=EOF) 35 { 36 sum=0; 37 for(int i=1;i<=n;i++) 38 { 39 scanf("%d",&a[i]); 40 sum+=a[i]; 41 } 42 solve(); 43 int t=0,ans[10000]; 44 for(int i=1;i<=sum;i++) 45 if(c1[i]==0) 46 { 47 t++; 48 ans[t]=i; 49 } 50 printf("%d ",t); 51 if(t){ 52 for(int i=1;i<t;i++) 53 printf("%d ",ans[i]); 54 printf("%d ",ans[t]); 55 } 56 } 57 return 0; 58 }