HDOJ 1709-The Balance【模拟】

小灰灰 2022-07-26 04:16 163阅读 0赞

# The Balance #

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

Total Submission(s): 7241 Accepted Submission(s): 2985

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 ][HDU 2007-Spring Programming Contest]

解题思路;

主要是一个天平有多个砝码问可以称出的重量，天平左右两边都可以放砝码。暴力就好，随便爆。

#include<stdio.h>
    #include<string.h>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int n;
    int num[30000];
    int ans[30000];
    int wc[30000];
    int pp[30000];
    
    int main()
    {
    	while(scanf("%d",&n)!=EOF)
    	{
    		memset(ans,0,sizeof(ans));
    		memset(pp,0,sizeof(pp));
    		int i,j;
    		int sum=0;
    		for(i=0;i<n;i++)
    		{
    			scanf("%d",&num[i]);
    			sum+=num[i];
    		}
    		for(i=0;i<n;i++)
    		{
    			for(j=1;j<=sum;j++)
    			{
    				if(ans[j]) 
    				{
    					pp[abs(j+num[i])]=1;
    					pp[abs(j-num[i])]=1;
    				}
    			}
    			for(j=1;j<=sum;j++)
    			{
    				if(ans[j]==0&&pp[j]==1) 
    				{
    					ans[j]=pp[j];
    				}
    			}
    			ans[num[i]]=1;
    			memset(pp,0,sizeof(pp));
    		}
    		int nn;
    		nn=0;
    		for(int k=1;k<=sum;k++)
    		{
    			if(!ans[k])
    			{
    				wc[nn++]=k;
    			}
    		}
    		printf("%d\n",nn);
    		for(int k=0;k<nn;k++)
    		{
    			if(k==0)
    			printf("%d",wc[k]);
    			else
    			printf(" %d",wc[k]);
    		}
    		if(nn!=0)
    		printf("\n");
    	}
    	return 0;
    }

[HDU 2007-Spring Programming Contest]: http://acm.hdu.edu.cn/search.php?field=problem&key=HDU+2007-Spring+Programming+Contest&source=1&searchmode=source