zoj 2795 Ambiguous permutations

來源:互聯網
上載者:User
//這題的大意為:給出一個數組,需要你驗證這個數組是否與自己的逆置換數組相等!//逆置換:例如給出數組1,4,3,2,而原來的數組順序為1, 2, 3, 4,就根據給出的數組作為原來數組排序的下標,//得出的逆置換為1, 4, 3, 2,與給出的數組相等,所以就為模糊排序!#include "iostream"#include "memory.h"using namespace std;int num[100010];int ans[100010];int main(){int n, i;while (cin >> n && n){bool flag = false;memset(num, 0, sizeof(num));memset(ans, 0, sizeof(ans));for (i = 1; i <= n; i++)cin >> num[i];for (i = 1; i <= n; i++)ans[num[i]] = i;for (i = 1; i <= n; i++){if (ans[i] != num[i]){flag = true;break;}}if (i == n + 1 && !flag)cout << "ambiguous" << endl;elsecout << "not ambiguous" << endl;}}

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