Balance

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.

2 4	-2 3 3 4 5 8

2

Romania OI 2002

 

题目的意思就是说将砝码挂在杠杆上，使得杠杆平衡，问你有多少重挂法?

 

dp[i][j]代表放第i个砝码的时候有j个重量,因为是要得到平衡时为0的重量，在计算时，重量可能小于0，所以我们估计出最大的重量是20*15*25=7500

所以将7500看成0，当j<0即为小于0时的状态

 

这样就可以得到状态方程

dp[i][j+w[k]*v[k]]+=dp[i-1][j]

 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 const int MAXN=20+5; 8 int w[MAXN],v[MAXN]; 9 int dp[MAXN][15005];10 int main()11 {12     int n,m;13     while(scanf("%d %d",&n,&m)!=EOF)14     {15         for(int i=1;i<=n;i++)16             scanf("%d",&w[i]);17         for(int i=1;i<=m;i++)18             scanf("%d",&v[i]);19             20         memset(dp,0,sizeof(dp));21         dp[0][7500]=1;22         for(int i=1;i<=m;i++)23         {24             for(int j=0;j<=15000;j++)25             {26                 if(dp[i-1][j])//如果为0，则代表之前没有得到这个状态，不用再计算了27                 {28                     for(int k=1;k<=n;k++)29                     dp[i][j+w[k]*v[i]]+=dp[i-1][j];30                 }31             }32         }33         printf("%d\n",dp[m][7500]);34     }35     return 0;36 }