POJ 1837: Balance DP...

POJ 1837: Balance DP...

 

Balance

Time Limit:1000 MS		Memory Limit:30000 K
Total Submissions:11878		Accepted:7417

 

Description

Gigel has a strange balance and he wants to poise it. Actually, the device is different from any other ordinary balance.
It orders two arms of negligible weight and each arm's length is 15. some hooks are attached to these arms and Gigel wants to hang up some weights from his collection of G weights (1 <= G <= 20) knowing that these weights have distinct values in the range 1 .. 25. gigel may droop any weight of any hook but he is forced to use all the weights.
Finally, Gigel managed to balance the device using the experience he gained at the National Olympus IAD in ICS. Now he wowould like to know in how many ways the device can be balanced.

Knowing the repartition of the hooks and the set of the weights write a program that calculates the number of possibilities to balance the device.
It is guaranteed that will exist at least one solution for each test case at the evaluation.

Input

The input has the following structure:
• The first line contains the number C (2 <= C <= 20) and the number G (2 <= G <= 20 );
• The next line contains C integer numbers (these numbers are also distinct and sorted in ascending order) in the range-15 .. 15 representing the repartition of the hooks; each number represents the position relative to the center of the balance on the X axis (when no weights are attached the device is balanced and lined up to the X axis; the absolute value of the distances represents the distance between the hook and the balance center and the sign of the numbers determines the arm of the balance to which the hook is attached: '-' for the left arm and '+' for the right arm );
• On the next line there are G natural, distinct and sorted in ascending order numbers in the range 1 .. 25 representing the weights 'values.

Output

The output contains the number M representing the number of possibilities to poise the balance.

Sample Input

2 4-2 3 3 4 5 8

Sample Output

2

The question is to give a scale, and then give you a few points to place the scale. The negative signs represent on the left of the scale, and exactly represent on the right of the scale.

Then I gave several shards and the weight of the shards. The question is how many balancing schemes are available with these sort operators.

After reading the question, I don't feel like this is a DP question... After reading this, I found that we should see such a small number and think of the enumeration status. In this way, we gradually push down the DP, so there are also a lot of questions, this is the idea of dividing 1, 2, 3, 4, and 6 stones.

This is also the case where DP [I] [j] is used to indicate that the status of the I-th percentile scale is j.

The scale status is calculated by 15*25*20, that is, up to 7500. -7500 to 7500. However, the array subscript must be greater than or equal to zero, so 0 ~ The value 15000 indicates.

DP [0] [7500] = 1 indicates that when a worker crashes, the status is in the balance state, and there is only one balance scheme.

And then... (I prefer to refer to it as enumeration ...) Push down.

Core code:

 

for(i=1;i<=G;i++){for(j=0;j<=15000;j++){for(k=1;k<=C;k++){dp[i][j] += dp[i-1][j-weight[i]*weizhi[k]];}}}

That is, the original state of each given worker is constantly enumerated, and its location is constantly enumerated, which leads to the State that can be reached this time.

 

Code:

 

# Include
 
  
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# Pragma warning (disable: 4996) using namespace std; // The largest state is that all weights are placed at the most endpoint. // 20*25*15 = 7500-7500 ~ 7500int dp [21] [15001]; int weizhi [21]; int weight [21]; int main () {memset (dp, 0, sizeof (dp )); dp [0] [7500] = 1; int C, G, I, j, k; cin> C> G; for (I = 1; I <= C; I ++) cin> weizhi [I]; for (I = 1; I <= G; I ++) cin> weight [I]; for (I = 1; I <= G; I ++) {for (j = 0; j <= 15000; j ++) {for (k = 1; k <= C; k ++) {dp [I] [j] + = dp [I-1] [j-weight [I] * weizhi [k];} cout <