Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)

以后每做完一场CF，解题报告都写在一起吧

 

暴力||二分 A - Bear and Elections

题意：有n个候选人，第一个候选人可以贿赂其他人拿到他们的票，问最少要贿赂多少张票第一个人才能赢

分析：正解竟然是暴力！没敢写暴力，卡了很久，导致这场比赛差点爆零！二分的话可以优化，但对于这题来说好像不需要。。。

收获：以后CF div2的A题果断暴力

 

代码(暴力)：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-30 0:40:46
* 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 = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n, x;

int main(void)    {
	scanf ("%d", &n);
	scanf ("%d", &a[0]);	n--;
	for (int i=1; i<=n; ++i)	scanf ("%d", &a[i]);
	int ans = 0;
	while (true)	{
		int mxi = 0;
		for (int i=1; i<=n; ++i)	{
			if (a[i] >= a[mxi])	mxi = i;
		}
		if (mxi == 0)	break;
		ans++;	a[0]++;	a[mxi]--;
	}
	printf ("%d\n", ans);

    return 0;
}

　　

代码(二分):

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-30 0:40:46
* 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 = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n;

bool cmp(int x, int y)	{
	return x > y;
}

bool check(int add)	{
	int y = a[0] + add;
	for (int i=1; i<=n; ++i)	{
		if (a[i] >= y)	{
			if (add <= 0)	return false;
			add -= (a[i] - (y - 1));
			if (add < 0)	return false;
		}
		else	break;
	}
	return true;
}

int main(void)    {
	scanf ("%d", &n);
	scanf ("%d", &a[0]);	n--;
	for (int i=1; i<=n; ++i)	scanf ("%d", &a[i]);
	sort (a+1, a+1+n, cmp);
	if (a[0] > a[1])	{
		puts ("0");	return 0;
	}
	int l = 0, r = 1000;
	while (l <= r)	{
		int mid = (l + r) >> 1;
		if (check (mid))	r = mid - 1;
		else l = mid + 1;
	}
	printf ("%d\n", l);

    return 0;
}

　　

暴力 B - Bear and Three Musketeers

题意：找一个三元环并且三个点的度数和-6最小

分析：三元环做过一题。这题能暴力跑过，DFS不用写。枚举边的两个端点，再找是否存在另外一个点构成三元环就可以了

收获：CF div2 B也暴力过

 

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-25 19:24:24
* File Name     :E_topo.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;
typedef pair<int, int> PII;
const int N = 4e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int deg[N];
bool g[N][N];
vector<PII> G;
int n, m;

int main(void)    {
	scanf ("%d%d", &n, &m);
	G.clear ();
    memset (g, false, sizeof (g));
    memset (deg, 0, sizeof (deg));

	for (int u, v, i=1; i<=m; ++i) {
		scanf ("%d%d", &u, &v);
        G.push_back (PII (u, v));
	    g[u][v] = true;    g[v][u] = true;
        deg[u]++;   deg[v]++;
    }

	int ans = INF;
    for (int i=0; i<G.size (); ++i) {
        int u = G[i].first, v = G[i].second;
        for (int j=1; j<=n; ++j)    {
            if (g[u][j] && g[v][j]) {
                ans = min (ans, deg[u] + deg[v] + deg[j] - 6);
            }
        }
    }

	printf ("%d\n", (ans == INF) ? -1 : ans);

    return 0;
}

　　

数学 C - Bear and Poker

题意：给n个数字，每个数字能*2或*3(可多次)，问是否能使得n个数相等

分析：要达到相等，那么每个数字除了2和3的数字外的剩余数字要相等，一个数除2或3是log级别的? 复杂度不会分析。。。

收获：其实这题很水，想到/2 /3就行了

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-30 1:42:08
* 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 = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];

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

int main(void)    {
	int n;	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	scanf ("%d", &a[i]);
	bool flag = true;
	for (int i=1; i<=n; ++i)	{
		while (a[i] % 2 == 0)	a[i] /= 2;
		while (a[i] % 3 == 0)	a[i] /= 3;
	}
	for (int i=1; i<n; ++i)	{
		if (a[i] != a[i+1])	{
			flag = false;	break;
		}
	}
	puts (flag ? "Yes" : "No");

    return 0;
}

　　

DP D - Bear and Blocks

题意：有n列高度不等的方块，每次可以将和空气(至少一面不和砖头或地面接触)接触的搬走，问要搬几次

分析：先考虑一列方块时只要搬一次。两列时，如果第二列高度大于第一列，那么搬两次，否则只搬一次。三列的情况同理

问题转换成最长连续上升序列，从前往后扫一次还有从后往前扫一次，两次取最小值，答案取最大值

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-30 16:56:38
* File Name     :D.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 = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N], dp1[N], dp2[N];

int main(void)    {
	int n;	scanf ("%d", &n);
	for (int i=0; i<n; ++i)	scanf ("%d", &a[i]);
	dp1[0] = 1;
	for (int i=1; i<n; ++i)	dp1[i] = min (a[i], dp1[i-1] + 1);
	dp2[n-1] = 1;
	for (int i=n-2; i>=0; --i)	dp2[i] = min (a[i], dp2[i+1] + 1);
	int ans = 0;
	for (int i=0; i<n; ++i)	{
		ans = max (ans, min (dp1[i], dp2[i]));
	}
	printf ("%d\n", ans);

    return 0;
}

　　

? E - Bear and Drawing