A - Balance

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A - Balance
POJ1837

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

题意：有一个天平，左臂右臂各长15，然后给出n，m，n代表有几个挂钩，挂钩给出负数代表在左臂的距离，正数则在右臂
　　　m代表有m个砝码（砝码重量 1<=g<=25），要你求出使得这个天平保持平衡有几种方法，要求所有砝码全部使用完
　　　　
思路：首先我们先要明确dp数组的作用，dp[i][j]中，i为放置的砝码数量，j为平衡状态，0为平衡,j<0左倾，j>0右倾，
由于j作为下标不能是负数，所以我们要找一个新的平衡点，

极端状态 ： 20 个 25g砝码 挂在 力臂位置15    20*25*15 = 7500   -7500~7500 ，

因为15*20*25 = 7500，所以平衡点设置为7500,
                                      然后我们可以得出动态方程 dp[i][j+w[i]*c[k])+=dp[i-1][j];

#include<cstdio>
#include<cstring>
int dp[25][16000] ; //  砝码数量 20  ， 挂上钩子后 左右相加会有一个平衡状态
int c[25] , w[25] ;

int main() {
    int n , m ;
    while(~scanf("%d%d" , &n , &m )) {

        for(int i=1 ; i<=n ; i++) {
            scanf("%d" , &c[i]) ; // 挂钩位置
        }
        for(int j=1 ; j<=m ; j++) {
            scanf("%d" , &w[j]) ; // 砝码重量
        }
        memset(dp , 0 , sizeof(dp)) ; // dp[i][j]平衡方法数量  表示放置 i 个砝码时的平衡状态 ， j == 7500 时平衡
        dp[0][7500] = 1  ; //7500为平衡状态    dp[0][7500]为放置 0 个砝码 平衡的 方法种类数
        // m 个砝码 n 个挂钩
        for(int i=1 ; i<=m ; i++) {
            for(int j=0 ; j<=15000 ; j++) {
                if(dp[i-1][j]) {
                    for(int k=1 ; k<=n ; k++) {
                        dp[i][j + c[k] * w[i]] += dp[i-1][j] ;
                    }
                }

            }
        }
        printf("%d\n" , dp[m][7500]) ;
    }
    return 0 ;
}

A - Balance

标签：each   input   poj   sam   class   enc   actual   actually   numbers   

原文地址：http://www.cnblogs.com/yi-ye-zhi-qiu/p/7719930.html