POJ1837Balance题解动态规划DP

Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3981		Accepted: 2300

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

据说是背包DP想了很久也没想明白

只有看foreverlin神牛的题解

关键是把a[i]*b[j]当做一个物品

题意：在-15到15这样一个轴上，有n个点，然后有m个砝码放到这n个点上去，使得其平衡，
即力臂和为0，注意所有的砝码都要用 典型的背包问题

状态：

d[i][j]表示用了i个砝码产生力臂和为j的方法数

由于j有负数加上7500偏移量

结果是d[m][7500]

状态转移方程：

d[i][k]+=d[i-1][k-b[i]*a[j]]

边界：

d[i][j]=0

d[1][a[i]*b[1]+7500]++

代码：

#include<cstdio> #include<cstring> int d[25][15005],a[25],b[25],n,m; int main() { while(scanf("%d%d",&n,&m)==2) { int i,j,k; memset(d,0,sizeof(d)); for(i=1;i<=n;i++) scanf("%d",a+i); for(i=1;i<=m;i++) scanf("%d",b+i); for(i=1;i<=n;i++) d[1][a[i]*b[1]+7500]++; for(i=2;i<=m;i++) for(j=1;j<=n;j++) for(k=15000-b[i]*a[j];k>0;k--) d[i][k]+=d[i-1][k-b[i]*a[j]]; printf("%d/n",d[m][7500]); } }