第八周训练a题 Balance

第八周训练a题 Balance

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

题意

有一个天平，臂长为15，给定c,为钩子个数，c[i]<0,代表在左边，c[i]>0代表在右边。给定g,为钩码的个数，g[i]为钩码质量。
要求有多少种方式使天平平衡，且砝码要全部用完。

思路

一眼题01背包。
一边能够达到的最大值 15*20*25=7500，左右两边，申请dp数组[25][15010]
因为左右两边都为7500，且左边用负数表示，所以的dp[i][7500]为平衡的点初始化为1；
例子分析一下：
先取第一个砝码，出现情况-6，9
接着取第二个砝码，出现情况-14，6，1，21
第三个砝码，出现情况-24，1，-4（-2*（3+5）+3*4），-9（-2*（4+5）+3*3），21（3*（4+5）+-2*3），16（3*（3+5）+-2*4），11（3*（3+4）+-2*5），36
最后一个砝码-40，0(-2*(3+4+5)+3*8)，…，0(-2*(4+8)+3*(3+5)),…,60

由上可知每种情况都可由上一种情况得到
推出dp动态方程为：dp[i][j+g[i]*c[k]]+=dp[i-1][j];

代码

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<queue>int dp[25][15010];int c[25],g[25];int main(){    int c,g;    while(scanf("%d%d",&c,&g)!=EOF){        memset(dp,0,sizeof(dp));        for(int i=1;i<=c;i++){            scanf("%d",&c[i]);        }        for(int i=1;i<=g;i++){            scanf("%d",&g[i]);        }        dp[0][7500]=1;                         //钩码不上的时候初始化为1        for(int i=1;i<=g;i++){            for(int j=0;j<=15000;j++){                if(dp[i-1][j]){                 //如果可以挂钩码                    for(int k=1;k<=c;k++){                        dp[i][j+g[i]*c[k]]+=dp[i-1][j];                    }                }            }        }        printf("%d\n",dp[g][7500]);    }    return 0;}