poj 1837 01背包

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Balance

Time Limit: 1000 MS Memory Limit: 30000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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DescriptionGigel 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.InputThe 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. OutputThe 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

/*01背包題意:C個鉤碼（2—20） G個物品（2—20） 鉤碼位置（-25—25） 物品重量（0—20）  物品都用上且天平平衡有多少種方案 dp[i][j]：掛前i個物品達到狀態j 狀態j的取值範圍時-25*25*20——25*25*20  所以j取（-7500--7500） 防止出現負值 所以令j==15000  即j==7500時為平衡位置想~~每次掛砝碼都會影響天平的平衡 即狀態j  影響因素是力臂=c[i]*w[k] (n,m影響它的取值）         掛前i個物品時狀態是dp[i-1][j] 則掛第i個物品後狀態變為dp[i][j+c[i]*w[k]]       假設dp[i-1][j]的值是num  那麼  dp[i][j+c[i]*w[k]]也是num      即dp[i][j+c[i]*w[k]]+=dp[i-1][j]   前面狀態影響後面的    */ #include <iostream> #include <string.h> #include <stdio.h> int dp[35][15001]; ///前i個物品達到j的狀態有的dp[][]種 int main() {     int n,m;  ///鉤子個數 砝碼個數     int c[35]; ///鉤子的位置     int w[35]; ///砝碼重量     scanf("%d%d",&n,&m);     for(int i=1;i<=n;i++)     scanf("%d",&c[i]);     for(int j=1;j<=m;j++)     scanf("%d",&w[j]);     memset(dp,0,sizeof(dp));       dp[0][7500]=1;   ///因為防止出現負數情況 所以dp[][1500]了  同時dp[][7500]是平衡狀態     for(int i=1;i<=m;i++)       {         for(int j=0;j<=15000;j++)           {             for(int k=1;k<=n;k++)               {                 dp[i][j+c[k]*w[i]]+=dp[i-1][j]; ///核心  在前面介紹             }         }     }     printf("%d\n",dp[m][7500]); }