HDU 5616 Jam's balance

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 806    Accepted Submission(s): 386


[align=left]Problem Description[/align]
Jim has a balance and N weights.
(1≤N≤20)

The balance can only tell whether things on different side are the same weight.

Weights can be put on left side or right side arbitrarily.

Please tell whether the balance can measure an object of weight M.
 

[align=left]Input[/align]
The first line is a integer
T(1≤T≤5),
means T test cases.

For each test case :

The first line is N,
means the number of weights.

The second line are N
number, i'th number wi(1≤wi≤100)
means the i'th weight's weight is wi.

The third line is a number M.
M
is the weight of the object being measured.
 

[align=left]Output[/align]
You should output the "YES"or"NO".
 

[align=left]Sample Input[/align]

1
2
1 4
3
2
4
5

 

[align=left]Sample Output[/align]

NO
YES
YES

Hint
For the Case 1:Put the 4 weight alone
For the Case 2:Put the 4 weight and 1 weight on both side

 

[align=left]Source[/align]
BestCoder Round #70

 

[align=left]Recommend[/align]
hujie   |   We have carefully selected several similar problems for you:  5654 5653 5652 5651 5650 
刚开始用DFS做的，没做出来，后来就用了DP，由于两边都可以放，所以和物品放相同的一边时相当于减，所以把01背包再反向进行一次。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[2100],wei[30],n;
int main()
{
int t,i,num,j,m;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
int sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&wei[i]);
sum+=wei[i];
}
dp[0]=1;
for(i=0;i<n;i++)
{
for(j=sum;j>=wei[i];j--)
{
dp[j]=max(dp[j],dp[j-wei[i]]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j+wei[i]<=sum;j++)
dp[j]=max(dp[j],dp[j+wei[i]]);
}
scanf("%d",&m);
while(m--)
{
scanf("%d",&num);
if(dp[num])
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}