codeforces 582A. GCD Table 解题报告

题目链接：http://codeforces.com/problemset/problem/582/A

　　网上很多题解，就不说了，直接贴代码= =

　　官方题解：

　　http://codeforces.com/blog/entry/20692

　　

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <map>
 #include <vector>
 using namespace std;

 const int maxn =  + ;
 map<int, int> cnt;
 vector<int> ans;
 int a[maxn*maxn];

 int GCD(int a, int b)
 {
     return b ==  ? a : GCD(b, a%b);
 }

 int main()
 {
     int n;
     #ifndef ONLINE_JUDGE
         freopen("in.txt", "r", stdin);
     #endif // ONLINE_JUDGE

     while (scanf("%d", &n) != EOF) {
         ans.clear();
         for (int i = ; i < n*n; i++) {
             scanf("%d", &a[i]);
             cnt[a[i]]++;
         }

         sort(a, a+n*n);
         for (int i = n*n-; i >= ; i--) {
             if (cnt[a[i]] <= ) {
                 continue;
             }
             cnt[a[i]]--;

             for (int j = ; j < ans.size(); j++) {
                 cnt[GCD(ans[j], a[i])] -= ;
             }
             ans.push_back(a[i]);
         }
         for (int i = ; i < ans.size(); i++) {
             cout << ans[i] << (i == ans.size()- ? '\n' : ' ');
         }
     }
     return ;
 }

官方解法代码（个人比较喜欢这个）

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 using namespace std;

 const int maxn =  + ;

 map<int, int> cnt;
 int ans[maxn];

 int gcd(int a, int b)
 {
     return b ==  ? a : gcd(b, a % b);
 }

 int main()
 {
     int a, n;

     #ifndef ONLINE_JUDGE
         freopen("in.txt", "r", stdin);
     #endif // ONLINE_JUDGE

     while (scanf("%d", &n) != EOF) {
         for (int i = ; i < n*n; i++) {
             scanf("%d", &a);
             cnt[-a]++;  // 为了将数组从大到小排序
         }

         int pos = n-;
         for (map<int, int>::iterator mp = cnt.begin(); mp != cnt.end(); ++mp) {
             int x = -mp->first;

             while (mp->second) {   // 当次数还有的时候
                 ans[pos] = x;
                 --mp->second;
                 for (int i = pos+; i < n; i++) {
                     cnt[-gcd(ans[i], x)] -= ;
                 }
                 pos--;
             }

         }
         for (int i = ; i < n; i++) {
             printf("%d%c", ans[i], (i == n- ? '\n' : ' '));
         }
     }
     return ;
 }