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
題意：有一個天平，左臂右臂各長15，然後給出n，m，n代表有幾個掛鉤，掛鉤給出負數代表在左臂的距離，正數則在右臂
m代表有m個砝碼，要你求出使得這個天平保持平衡有幾種方法，要求所有砝碼全部使用完
思路：首先我們先要明確dp數組的作用，dp[i][j]中，i為放置的砝碼數量，j為平衡狀態，0為平衡,j<0左傾，j>0右傾，由於j作為下標不能是負數，所以我們要找一個新的平衡點，因為15*20*20 = 7500，所以平衡點設置為7500,
然後我們可以得出動態方程 dp[i][j+w[i]*c[k])+=dp[i-1][j];
#include
#include
#include
#include
using namespace std;
int dp[25][16000];
int c[25],w[25];
int main()
{
int n,m,i,j,k;
while(~scanf("%d%d",&n,&m))
{
for(i = 1; i<=n; i++)
scanf("%d",&c[i]);
for(i = 1; i<=m; i++)
scanf("%d",&w[i]);
memset(dp,0,sizeof(dp));
dp[0][7500] = 1;
for(i = 1; i<=m; i++)
{
for(j = 0; j<=15000; j++)
if(dp[i-1][j])//一個小小的優化
for(k = 1; k<=n; k++)
dp[i][j+c[k]*w[i]]+=dp[i-1][j];
}
printf("%d\n",dp[m][7500]);
}
return 0;
}
