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 = 500 + 5;
map<int, int> cnt;
vector<int> ans;
int a[maxn*maxn];

int GCD(int a, int b)
{
    return b == 0 ? 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 = 0; i < n*n; i++) {
            scanf("%d", &a[i]);
            cnt[a[i]]++;
        }

        sort(a, a+n*n);
        for (int i = n*n-1; i >= 0; i--) {
            if (cnt[a[i]] <= 0) {
                continue;
            }
            cnt[a[i]]--;

            for (int j = 0; j < ans.size(); j++) {
                cnt[GCD(ans[j], a[i])] -= 2;
            }
            ans.push_back(a[i]);
        }
        for (int i = 0; i < ans.size(); i++) {
            cout << ans[i] << (i == ans.size()-1 ? '
' : ' ');
        }
    }
    return 0;
}

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 = 500 + 5;

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

int gcd(int a, int b)
{
    return b == 0 ? 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 = 0; i < n*n; i++) {
            scanf("%d", &a);
            cnt[-a]++;  // 为了将数组从大到小排序
        }

        int pos = n-1;
        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+1; i < n; i++) {
                    cnt[-gcd(ans[i], x)] -= 2;
                }
                pos--;
            }

        }
        for (int i = 0; i < n; i++) {
            printf("%d%c", ans[i], (i == n-1 ? '
' : ' '));
        }
    }
    return 0;
}