Codeforces Round #323 (Div. 2)

被进爷坑了，第二天的比赛改到了12点

 

水 A - Asphalting Roads

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/3 星期六 21:53:09
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 55;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
bool r[N], c[N];

int main(void)    {
	memset (r, false, sizeof (r));
	memset (c, false, sizeof (c));
	int n;	scanf ("%d", &n);
	vector<int> ans;
    n = n * n;
	for (int x, y, i=1; i<=n; ++i)	{
		scanf ("%d%d", &x, &y);
		if (!r[x] && !c[y])	{
			r[x] = c[y] = true;
			ans.push_back (i);
		}
	}
	for (int i=0; i<ans.size (); ++i)	{
		printf ("%d%c", ans[i], (i == ans.size () - 1) ? '\n' : ' ');
	}

    return 0;
}

 

水 B - Robot's Task

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/3 星期六 21:53:24
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N];

int main(void)    {
    int n;  scanf ("%d", &n);
    for (int i=1; i<=n; ++i)    {
        scanf ("%d", &a[i]);
    }
    int ans = 0, d = 1, m = 0, p = 0;
    bool flag = false;
    while (true)   {
        if (d == 1) {
            for (int i=p+1; i<=n; ++i)    {
                if (m >= a[i])  {
                    a[i] = INF;
                    m++;    p = i;
                    flag = true;
                }
            }
            if (m == n) break;
            d ^= 1; ans++;
        }
        else    {
            for (int i=p-1; i>=1; --i)    {
                if (m >= a[i])  {
                    a[i] = INF;
                    m++;    p = i;
                    flag = true;
                }
            }
            if (m == n) break;
            d ^= 1; ans++;
        }
    }
    printf ("%d\n", ans);

    return 0;
}

 

贪心 C - GCD Table

题意：给了一张GCD表，问原来的求GCD的那些数

分析：从大到小找，最大的数和其他的数的GCD都不大于它，每次找到一个就能把它和已知的答案的GCD给删除，map+暴力就可以了

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/3 星期六 21:53:35
* File Name     :C.cpp
 ************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 5e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N*N];
int ans[N];
map<int, int> cnt;

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

int main(void)    {
    int n;  scanf ("%d", &n);
    int m = n;
    n = n * n;
    for (int i=1; i<=n; ++i)    {
        scanf ("%d", &a[i]);
        cnt[-a[i]]++;
    }
    int pos = m;
    map<int, int>::iterator it;
    for (it=cnt.begin (); it!=cnt.end (); ++it) {
        int x = -it -> first;
        while (it -> second)    {
            ans[pos] = x;
            --it -> second;
            for (int i=pos+1; i<=m; ++i)    {
                cnt[-GCD (ans[pos], ans[i])] -= 2;
            }
            pos--;
        }
    }

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

    return 0;
}