在C ++中执行给定操作后,数组中的最大可能乘积
在本教程中,我们将讨论执行给定操作后在阵列中找到最大可能乘积的程序
为此,我们将得到一个大小为N的数组。我们的任务是执行N-1操作(更改a[j]→a[i]*a[j]并删除a[i]值,或者只是删除该值a[i](仅一次)的最大值),以使剩余值仅是最大值。
示例
#include <bits/stdc++.h> using namespace std; //打印业务 void MaximumProduct(int a[], int n) { int cntneg = 0; int cntzero = 0; int used[n] = { 0 }; int pos = -1; for (int i = 0; i < n; ++i) { if (a[i] == 0) { used[i] = 1; cntzero++; } if (a[i] < 0) { cntneg++; if (pos == -1 || abs(a[pos]) > abs(a[i])) pos = i; } } if (cntneg % 2 == 1) used[pos] = 1; if (cntzero == n || (cntzero == n - 1 && cntneg == 1)) { for (int i = 0; i < n - 1; ++i) cout << 1 << " " << i + 1 << " " << i + 2 << endl; return; } int lst = -1; for (int i = 0; i < n; ++i) { if (used[i]) { if (lst != -1) cout << 1 << " " << lst + 1 << " " << i + 1 << endl; lst = i; } } if (lst != -1) cout << 2 << " " << lst + 1 << endl; lst = -1; for (int i = 0; i < n; ++i) { if (!used[i]) { if (lst != -1) cout << 1 << " " << lst + 1 << " " << i + 1 << endl; lst = i; } } } int main() { int a[] = { 5, -2, 0, 1, -3 }; int n = sizeof(a) / sizeof(a[0]); MaximumProduct(a, n); return 0; }
输出结果
2 3 1 1 2 1 2 4 1 4 5