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 eva luation. 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 Source Romania OI 2002 |
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思路分析：
应该是二维数组，dp[i][j] 表示前i个称出来j状态的数量数，j表示重量，小于7500 表示左边重，反之右边重。
在此可以用滚动数组。用一个tmp记录第i个砝码扩展出来的状态，然后最后赋给dp
在这里要说一下，每一次dp都要memset的原因
我自己纠结了好久，因为之前的每一次状态都是上“一”个砝码扩展出来的，而且他是要全部都用到。所以比如说
第一个砝码扩展出来了
-3 3 这两个状态，
第二个砝码得到了转移 6
那么 当-3 + 6 之后 -3 这个状态就不能要了，因为这样的话只用了一个砝码。
16MS
#include#include #include #include using namespace std; int dp[17000]; int tmp[17000]; int main() { int c[25]; int g[25]; int C,N; while(scanf("%d%d",&C,&N)!=EOF) { memset(dp,0,sizeof dp); for(int i=1;i<=C;i++) scanf("%d",&c[i]); for(int i=1;i<=N;i++) scanf("%d",&g[i]); dp[7500]=1; for(int i=1;i<=N;i++) { memset(tmp,0,sizeof(tmp)); for(int k=0;k<=15000;k++) if(dp[k]) for(int j=1;j<=C;j++) { int v=g[i]*c[j]; tmp[k+v]+=dp[k]; } memset(dp,0,sizeof dp); for(int j=0;j<=15000;j++) if(tmp[j])dp[j]+=tmp[j]; } printf("%d\n",tmp[7500]); } return 0; }