74 lines
1.3 KiB
C++
74 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 30;
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int a[N];
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int n;
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//最终答案
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int res;
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//互质
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bool isHz(int a, int b) {
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for (int i = 2; i <= min(a, b); i++) //a,b可能存在的公因子
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if (a % i == 0 && b % i == 0)
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return false;
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return true;
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}
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bool found = false; //最终是不是能表示出来 found=true是可以表达，found=false是不能表达
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//能表示
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void dfs(int sum, int u) { //用a[]中所有元素组合，能不能表示出x这个数字，如果能返回true,否则返回false
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if (u == n + 1) {
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if (sum == 0)
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found = true;
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return;
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}
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//尝试当前a[u]选择i个
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for (int i = 0; sum >= i * a[u]; i++) //i是枚举的a[u]这个数的个数
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dfs(sum - i * a[u], u + 1);
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}
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++)
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cin >> a[i];
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//排一下序，由小到大
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sort(a + 1, a + 1 + n);
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//a数组中，是不是存在两个互质的数
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bool flag = false;
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int mx;//疑似上线
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for (int i = 1; i <= n; i++) {
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for (int j = i + 1; j <= n; j++)
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if (isHz(a[i], a[j])) {
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flag = true;
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mx = a[i] * a[j] - a[i] - a[j];
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break;
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}
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if (flag)
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break;
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}
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if (flag) {
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for (int i = 1; i <= mx; i++) {
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//一定要记得清空
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found = false;
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dfs(i, 1);
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if (!found)
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res++;
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}
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cout << res << endl;
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} else
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cout << -1 << endl;
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return 0;
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}
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