【HDU】5616 - Jam's balance（母函数变型）
【HDU】5616 - Jam's balance（母函数变型）
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Jam's balance
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1133 Accepted Submission(s): 531
Problem Description
Jim has a balance and N weights. 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 , means T test cases. For each test case : The first line is , means the number of weights. The second line are number, i'th number means the i'th weight's weight is . The third line is a number . 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
5Sample 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 sideSource
BestCoder Round #70
同样是母函数，但是砝码可以放到两边，计数的时候算一下 abs( j - k ) 就行了。
代码如下：
#include <cstdio>
#include <cstring>
int c[2000+11];
int t[2000+11];
int sum;
int abs(int x)
{
return x < 0 ? -x : x;
}
void init()
{
sum = 0;
memset (c,0,sizeof (c));
memset (t,0,sizeof (t));
}
int main()
{
int u;
int w[22];
int n,q;
scanf ("%d",&u);
while (u--)
{
scanf ("%d",&n);
init(); //初始化
for (int i = 1 ; i <= n ; i++)
{
scanf ("%d",&w[i]);
sum += w[i];
}
c[w[1]] = c[0] = 1;
for (int i = 2 ; i <= n ; i++)
{
for (int j = 0 ; j <= sum ; j++)
for (int k = 0 ; k <= w[i] && k+j <= sum ; k += w[i])
{
t[k+j] += c[j];
t[abs(k-j)] += c[j]; //砝码放到另一个盘子上
}
for (int j = 0 ; j <= sum ; j++)
{
c[j] = t[j];
t[j] = 0;
}
}
scanf ("%d",&q);
while (q--)
{
int x;
scanf ("%d",&x);
printf (c[x] ? "YES\n" : "NO\n");
}
}
return 0;
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