Balance
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 13761 | Accepted: 8652 |
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.
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.
• 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
题解：
题目大意：
有一个天平，天平左右两边各有若干个钩子，总共有C个钩子，有G个钩码，求将钩码全部挂到钩子上使天平平衡的方法的总数。
其中可以把天枰看做一个以x轴0点作为平衡点的横轴
输入：
2 4 //C 钩子数 与 G钩码数
-2 3 //负数：左边的钩子距离天平中央的距离；正数：右边的钩子距离天平中央的距离c[k]
3 4 5 8 //G个重物的质量w[i]
分析：
对于这个题我们首先想到的是穷举暴搜，然而对于O(20^20)显然会TLE，而且会T的很彻底
所以怎么办呢？？？
我们再来看一遍题：
它说使得钩码全部挂到钩子上的全部方案，那么当前的方案一定会是从前面放较少的钩码得到，所以就变成了一个DP题。
那么怎么转移呢？？？
我们首先定义一个状态：dp[i][j]表示放前i个钩码时天平的平衡度为j，平衡度表示天平两端的重量差，所以最大的平衡度为25*20*15=7500，就是把所有钩码放到一边的最大长度。因为天平有左右，平衡度最大时为15*20*25，为7500。因为钩码在左边时，我们定义的为负，所以就将数组加7500，所以当平衡度为7500时恰好平衡，那么我们就成了求dp[g][7500]了。
对于转移，我们发现，当前你要挂一个钩码的方案数一定是没挂这个钩码时的总方案数，所以方程为dp[i][j+w[i]*c[i]] = ∑ dp[i-1][j]
初始化时：dp[0][7500] = 1;
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[25][15005];
int c[25];
int w[25];
int main()
{
int n;
int g;
cin >> n >> g;
for (int i = 1; i <= n; i++)
scanf("%d", &c[i]);
for (int i = 1; i <= g; i++)
scanf("%d", &w[i]);
memset(dp, 0, sizeof(dp));
dp[0][7500] = 1;
for (int i = 1; i <= g; i++)
{
for (int j = 0; j <= 15000; j++)
{
if (dp[i - 1][j])
for (int k = 1; k <= n; k++)
{
dp[i][j + w[i] * c[k]] += dp[i - 1][j];
}
}
}
cout << dp[g][7500];
return 0;
}