poj 1837 Balance

Balance

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 10241 Accepted: 6330

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 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. 

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

Source

Romania OI 2002

真心觉得自己弱爆了，原本很简单的一道DP自己看了3个小时，才有自己的一些算法，可能是我第一次做这个DP的题吧。总之就一个字弱。然后小优的 解题报告真心不错上面解释的很详细。我只是根据他的状态方程，把程序写出来而已。题意的话你们看他的吧小优博客。我就附上我看了DP方程后的代码吧，希望不要喷我，我很水

/*#include<iostream>#include<cstring>using namespace std;int dp[21][15001];//挂的方法数int main(){int i,j,k;int n;//n为沟的个数int g;// 为挂的砝码数；int c[21];//为他的距离int w[21];//为重量；cin>>n>>g;for(i=1;i<=n;i++)cin>>c[i];//输入沟的距离 for(i=1;i<=g;i++)cin>>w[i];//输入第i个的重量 memset(dp,0,sizeof(dp));dp[0][7500]=1;for(i=1;i<=g;i++)for(j=0;j<=15000;j++)if(dp[i-1][j])for(k=1;k<=n;k++)dp[i][j+w[i]*c[k]]+=dp[i-1][j];cout<<dp[g][7500]<<endl;return 0;} */#include<iostream>#include<cstring>using namespace std;int dp[21][150001];//这个也是小优给的int main(){int i,j,k;int c[21];//距离int w[21];//每个砝码的重量int n;//钩子的数量 int p;//砝码的数量 cin>>n>>p;for(i=1;i<=n;i++)cin>>c[i];for(j=1;j<=p;j++)cin>>w[j];memset(dp,0,sizeof(dp));dp[0][7500]=1;for(k=1;k<=p;k++)for(j=0;j<=15000;j++)if(dp[k-1][j])for(i=1;i<=n;i++)dp[k][c[i]*w[k]+j]+=dp[k-1][j];cout<<dp[p][7500]<<endl;return 0;} 

不解释啊，感觉动态规划博大精深