題干：

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

題目大意：

? 有一個天平，左側(cè)15的長度，右側(cè)15的長度。上面有C個地方可以放砝碼，分別告訴你x坐標（天平軸是坐標零點）。然后告訴你有G個砝碼，問你可以使得天平平衡的總方案數(shù)。

解題報告：

? dp[i][j]代表放置了前i個砝碼，右側(cè)比左側(cè)重量多j，的方案數(shù)。然后xjb遞推就行了。本來是“人人為我型”T了，然后改成“我為人人型”并且加了點優(yōu)化才A。。（好像是因為上界開太大了所以T了⑧）

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; ll n; int a[MAX],b[MAX]; int dp[33][MAX];//代表右邊減左邊的 const int zero = 1e5; int main() {int c,g;while(~scanf("%d%d",&c,&g)) {memset(dp,0,sizeof dp);dp[0][zero] = 1;for(int i = 1; i<=c; i++) scanf("%d",a+i);for(int i = 1; i<=g; i++) scanf("%d",b+i);for(int i = 1; i<=g; i++) {for(int j = 0; j<MAX; j++) {if(dp[i-1][j] == 0) continue;for(int k = 1; k<=c; k++) {if(j+a[k]*b[i] >= 0 && j+a[k]*b[i] < MAX) dp[i][j+a[k]*b[i]] += dp[i-1][j]; }}}printf("%d\n",dp[g][zero]);}return 0 ; }

?

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