题目传送门
平衡问题，把每个砝码在每个位置的权值算出来，每个砝码一个分组，几个位置几个物品，最后求的是价值和为0的方案数。
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C个挂钩，G个砝码，求使天平平衡的方案数。
分析：
- 每个砝码看做一组，同一砝码在不同位置的贡献作为组内不同物品
- 方案数问题，状态转移方程：dp[k][j +权值] += dp[k - 1][j];
AC代码：
#include <iostream>
#include <vector>
#include <utility>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <cstdio>
#include <fstream>
#include <set>
using namespace std;
typedef long long ll;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define INF 0x3f3f3fconst int MAXN = 7500 * 2 + 7;int dp[25][MAXN];
int c[25], g[25];
vector<int>vec[MAXN];
int main() {
ios;int C, G;int balance = 7500;while (cin >> C >> G) {
memset(dp, 0, sizeof dp);for (int i = 0; i <= C * G; i++) {
vec[i].clear();}dp[0][balance] = 1;for (int i = 1; i <= C; i++)cin >> c[i];for (int i = 1; i <= G; i++) {
cin >> g[i];for (int j = 1; j <= C; j++) {
vec[i].push_back(g[i] * c[j]);}}for (int k = 1; k <= G; k++) {
for (int j = 15000; j >= 0; j--) {
for (int i = 0; i < vec[k].size(); i++) {
dp[k][j + vec[k][i]] += dp[k - 1][j];}}}cout << dp[G][balance] << endl;}return 0;
}