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題目：http://poj.org/problem?id=1837

Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16411		Accepted: 10310

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


題目意思是給你一盞天平，天平左右兩邊的不同位置有鉤子，給定一定數量的砝碼，將所有砝碼掛在天平上，問有多少種掛法？
本題用動態規劃，分析：
　　　　首先：定義一個數組dp[i][j] : 表示用前i個砝碼，掛在天平上，使得天平達到狀態j（狀態j是指：j<0 天平左邊比右邊的重j克；j>0 天平右邊比左邊重j克；) 的掛法總數；
　　　　　　　weight[i]：表示第i個砝碼的重量；
　　　　　　　dis[k] :表示第k個鉤子的位置;
　　　　　　　hook:表示鉤子的數量
　　　　　　　　
　　　　其次：根據題目意思可知，在極端情況，給定20個砝碼，　每個均為25克，在天平左邊 15 厘米 處 有鉤子，將所有砝碼放在天平最左邊15厘米處，此時 j= - 20*25*15=7500　，
　　　　　　　由於j作為數組下標不能是負數，所以我們可以在編寫代碼時將j右移7500個單位，平衡點j=0變為j=7500
　　　　
　　　　所以：其狀態轉移方程為
　　　　　　　　dp[i][j] += dp[i-1][j-weight[i]*dis[k] ] 其中 k=1...hook; 
代碼實現
 
：
　　　

#include<iostream>
#include<cstring>
using namespace std;
int main(){
	int G,C;
	int hook[25];
	while(cin>>C>>G){
		for(int i=1;i<=C;i++){
			scanf("%d",&hook[i]);
		}
		int a[21][15005];
		memset(a,0,sizeof(a));
		int we1;
		scanf("%d",&we1);
		for(int i=1;i<=C;i++){
			a[1][7500+hook[i]*we1]=1;
		}
	//	cout<<a[1][7494]<<" "<<a[1][7509]<<endl;
		for(int i=2;i<=G;i++){
			int x;
			scanf("%d",&x);
			
			for(int j=1;j<=15000;j++){
				for(int k=1;k<=C;k++){
					if(j-x*hook[k]>0&&j-x*hook[k]<=15000){
						a[i][j]+=a[i-1][j-x*hook[k]]; 		
					//	if(a[i][j]) cout<<a[i][j]<<" "<<j-x*hook[k]<<" "<<i<<" "<<j<<endl;
					}		
				}
			}
			
		}
	//	for(int i=1;i<=20;i++)
		cout<<a[G][7500]<<" "<<endl;
	}			
}

　　

動態規劃dp