HDOJ 1709-The Balance【模拟】

The Balance

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

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

[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

[align=left]Source[/align]
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;
}