POJ 1837-Balance

POJ 1837-Balance

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 10845 Accepted: 6742

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 Olympiad in Informatics. Now he would 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.

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.

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

Romania OI 2002


后来看了某前辈的解题报告,他提出了一个平衡度的概念,并推导出如下的状态转换方程:①dp[i][ j+ w[i]*c[k] ]= ∑(dp[i-1][j])。dp[i][j]表示前i个砝码使平衡度为j共有多少种方案数。

按照常理,我们得到的状态转换方程为:②dp[i][j] =∑(dp[i - 1][j - c[i] * w[i]]),表示dp[i][j]等于前i-1个砝码使平衡度为j-x的方案数之和,这个x就是第i个砝码挂在c个挂钩中的某一个产生的力臂。稍加推导就能得出①等价于②。


using namespace std;
int c[21];
int w[21];
int dp[21][15001];

int main()
     int n,g;
     for(int i=1;i<=n;i++)
     for(int i=1;i<=g;i++)

     for(int i=1;i<=g;i++)//对于每一个砝码
          for(int j=0;j<=15000;j++)//挂上去之后可能使天平出现0-15000中的任意一种状态
                    for(int k=1;k<=n;k++)
     return 0;


One thought on “POJ 1837-Balance

  1. Pingback: 从一个数字序列中取若干个数字使得和为某个数,问共有多少种取数方案? | bitJoy

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