hdu 5616(dp)

Jam's balance

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

Total Submission(s): 733 Accepted Submission(s): 351



Problem Description

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.

Input

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.

Output

You should output the "YES"or"NO".

Sample Input

1
2
1 4
3
2
4
5

Sample Output

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

Source

BestCoder Round #70

//好久没刷题了。。这题求的是给你N个数是否可以通过加减运算得到另外一个数

//主要利用类01背包的思想 dp[i][k] 为前i个数是否可以得到k 可以得到为1不能得到为0

//dp[i][k] = max(dp[i-1][k],dp[i-1][abs(k-a[i])]); dp[i][k] =max(dp[i][k],dp[i-1][k+a[i]])

#include <bits/stdc++.h>
using namespace std;
const int N=25;
int arr
;
int dp
[N*100];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
for(int i=0;i<=n;i++)
for(int k=0;k<N*100;k++)
dp[i][k]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",arr+i);
dp[i][0]=1;
}
dp[0][0]=1;
for(int i=1;i<=n;i++)
{
for(int k=1;k<n*100;k++)
{
dp[i][k]=max(dp[i-1][k],dp[i-1][abs(k-arr[i])]);
dp[i][k]=max(dp[i][k],dp[i-1][k+arr[i]]);
}
}
int m;
scanf("%d",&m);
while(m--)
{
int val;
scanf("%d",&val);
if(dp
[val])
puts("YES");
else
puts("NO");
}
}

return 0;
}