The Balance

The Balance

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

Total Submission(s): 135 Accepted Submission(s): 88

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

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

/*
题意：给出n个砝码的质量，让你求出在[1,s]范围内不能组成的质量，s是质量总和，这个题不一样的地方是天平两边都可以放砝码

初步思路：可以看成一个背包问题，每个取或不取，但是这个题有一个新的状态，就是
*/
#include<bits/stdc++.h>
using namespace std;
int v[];
int dp[];//dp[i]表示i金额能组成或不能组成
int vis[];//记录一下能组成的金额
int n;
int s=;
void init(){
    memset(vis,,sizeof vis);
    memset(dp,,sizeof dp);
    s=;
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=;i<=n;i++){
            scanf("%d",&v[i]);
            s+=v[i];
        }
        vis[]=;
        for(int i=;i<=n;i++){
            memset(dp,,sizeof dp);
            for(int j=;j<=s;j++){
                if(vis[j]){//j金额能组成
                    dp[j+v[i]]=;//加上当前金额那么也是可以的
                    dp[abs(j-v[i])]=;//减去这个金额的也是可以的
                }
            }
            for(int j=;j<=s;j++){
                if(dp[j]) vis[j]=;
            }
        }
        int res=;
        for(int i=;i<=s;i++)
            if(!vis[i])
                res++;
        printf("%d\n",res);
        if(res==) continue;
        int f=;
        for(int i=;i<=s;i++){
            if(!vis[i]){
                if(f==){
                    printf("%d",i);
                    f=;
                }else{
                    printf(" %d",i);
                }
            }
        }
        printf("\n");
    }
    return ;

}