Codeforces Beta Round #97 (Div. 2) 【完整题解】

KIDx 的解题报告
题目链接：http://codeforces.com/contest/136
以下省略头文件

前三题是超级水题，不解释；后两题是很不错的水题，详细解释

 

A题

#include <iostream>
using namespace std;
#define M 105
int pre[M];

int main()
{
	int n, i, x;
	while (~scanf ("%d", &n))
	{
		for (i = 1; i <= n; i++)
			scanf ("%d", &x), pre[x] = i;
		for (i = 1; i < n; i++)
			printf ("%d ", pre[i]);
		printf ("%d\n", pre[i]);
	}
    return 0;
}

 

B题

#include <iostream>
#include <cmath>
using namespace std;
#define M 105

int a[M], c[M], b[M];

int main()
{
	int x, y, k, n, maxs, res, i;
	while (~scanf ("%d%d", &x, &y))
	{
		memset (a, 0, sizeof(a));
		memset (c, 0, sizeof(c));
		k = 0;
		while (x)
		{
			a[k++] = x % 3;
			x /= 3;
		}
		n = 0;
		while (y)
		{
			c[n++] = y % 3;
			y /= 3;
		}
		maxs = max (n, k);
		res = 0;
		for (i = 0; i < maxs; i++)
		{
			if (c[i] >= a[i])
				b[i] = c[i] - a[i];
			else b[i] = c[i] + 3 - a[i];
			res += b[i] * pow (3.0, i);
		}
		printf ("%d\n", res);
	}
    return 0;
}

 

C题

#include <iostream>
#include <algorithm>
using namespace std;
#define M 100005
int a[M];

int main()
{
	int n, i;
	while (~scanf ("%d", &n))
	{
		for (i = 0; i < n; i++)
			scanf ("%d", a+i);
		sort (a, a+n);
		if (a[n-1] == 1)    //注意一下全部是1的情况即可
			a[n-1] = 2;
		else a[n-1] = 1, sort (a, a+n);
		printf ("%d", a[0]);
		for (i = 1; i < n; i++)
			printf (" %d", a[i]);
		printf ("\n");
	}
    return 0;
}

 

D题

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define eps 1e-8
#define M 10

struct point{
	double x, y;
}p[M], tp[M];

double dist (point a, point b)
{
	return sqrt ((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

double dianmulti (point a, point b, point c)    //求ab与bc的点积
{
	double x1 = b.x - a.x;
	double y1 = b.y - a.y;
	double x2 = c.x - b.x;
	double y2 = c.y - b.y;
	return x1*x2+y1*y2;
}

int main()
{
	int i, j, k, l, ind, q, v, ed, mid, snum[M], ans;
	bool hash[M];
	double maxs, a[5];
	while (~scanf ("%lf%lf", &p[0].x, &p[0].y))
	{
		for (i = 1; i < 8; i++)
			scanf ("%lf%lf", &p[i].x, &p[i].y);
		memset (hash, false, sizeof(hash));
		for (i = 0; i < 8; i++)
		{
			for (j = i + 1; j < 8; j++)
			{
				for (k = j + 1; k < 8; k++)
				{
					for (l = k + 1; l < 8; l++)
					{
						ind = ans = 0;
						
						tp[ind++] = p[i];
						tp[ind++] = p[j];
						tp[ind++] = p[k];
						tp[ind++] = p[l];		//tp存放四边形的点
						snum[ans++] = i + 1;
						snum[ans++] = j + 1;
						snum[ans++] = k + 1;
						snum[ans++] = l + 1;    //记录组成四边形的点的编号

						maxs = 0;
						for (q = 1; q < ind; q++)
						{	//找出v使得tp[0],tp[v]组成对角线
							double dis = dist (tp[0], tp[q]);
							if (dis > maxs)
								maxs = dis, v = q;
						}
						ed = 0;
						for (q = 1; q < ind; q++)
						{
							if (q != v)
							{	//找出四边形边长
								a[ed++] = dist (tp[0], tp[q]);
								a[ed++] = dist (tp[v], tp[q]);
							}
						}

						if (fabs (a[0]-a[1]) < eps &&
							fabs (a[0]-a[2]) < eps &&
							fabs (a[0]-a[3]) < eps)
						{	//判正方形四条边相等
							for (q = 1; q < ind; q++)
							{
								if (q != v)
								{
									mid = q;	//tp[mid]是不同于tp[0]，tp[v]的点
									break;
								}
							}
							if (fabs (dianmulti (tp[0], tp[mid], tp[v])) < eps)
							{	//点积判0->mid是否垂直于mid->v
								//来到这里，说明i，j，k，l可以组成正方形
								//以下基本上同理了！
								ind = 0;
								for (q = 0; q < 8; q++)
								{	//找到非正方形的点存放到tp
									if (q != i && q != j && q != k && q != l)
										tp[ind++] = p[q];
								}
								maxs = 0;
								for (q = 1; q < ind; q++)
								{
									double dis = dist (tp[0], tp[q]);
									if (dis > maxs)
										maxs = dis, v = q;
								}
								ed = 0;
								for (q = 1; q < ind; q++)
								{
									if (q != v)
									{
										a[ed++] = dist (tp[0], tp[q]);
										a[ed++] = dist (tp[v], tp[q]);
									}
								}
								sort (a, a+ed);    //排序方便判对边相等
								if (fabs (a[0]-a[1]) < eps &&
									fabs (a[2]-a[3]) < eps)
								{	//判长方形对边相等
									for (q = 1; q < ind; q++)
									{
										if (q != v)
										{
											mid = q;
											break;
										}
									}
									if (fabs (dianmulti (tp[0],
										tp[mid], tp[v])) < eps)
										goto end;//到这里,说明另外4点可以组成长方形
								}
							}
						}
					}
				}
			}
		}
		puts ("NO");
		continue;
end:;
		puts ("YES");
		for (i = 0; i < ans - 1; i++)
			printf ("%d ", snum[i]), hash[snum[i]] = true;
		printf ("%d\n", snum[i]), hash[snum[i]] = true;
		ans = 0;
		for (i = 1; i <= 8; i++)
		{
			if (!hash[i])
				snum[ans++] = i;		//找出不是组成正方形的点
		}
		for (i = 0; i < ans - 1; i++)	//再次输出
			printf ("%d ", snum[i]);
		printf ("%d\n", snum[i]);
	}
    return 0;
}

 

E题

思路：

设0的个数为n0，1的个数为n1，遇到？也会有n0++，n1++，因为0,1的个数完全可能由？构成

设w为？的个数

因为答案只有4种情况(00,01,10,11)，只要分别想办法试着构造即可

①n0 > len-n0 (0的个数>1的个数)即必然可以构造出"00"

②n1 > len-n1+1 (1的个数>0的个数+1)必然可以构造出"11"

③除了①②情况外只剩下两种情况：

1.a个1,a个0(如果n0<=n1&&n0>=len/2则有可能出现此情况)：

按照游戏程序，必然要删掉a-1个1，a-1个0

2.a+1个1,a个0(n0>n1&&n1>=(len+1)/2……)：

按照游戏程序，必然要删掉a个1，a-1个0

即最后必然只剩下一个1，一个0

所以最后一个字符显然是不会被删掉的

那么

如果最后一个字符是1，显然答案01是可构造的

如果最后一个字符是0，显然10是可构造的

如果最后一个字符是？，则需分类讨论：

1.设？=1，则对于当前n0--，n1不变，然后再用上面绿色条件判定

2.设？=0，则对于当前n1--，n0不变，然后同理

#include <iostream>
using namespace std;
#define M 100005

char s[M];

int main()
{
	int i, len, n0, n1, w;
	while (~scanf ("%s", s))
	{
		len = strlen (s);
		n0 = n1 = w = 0;
		for (i = 0; i < len; i++)
		{
			if (s[i] == '?')
				n0++, n1++, w++;
			if (s[i] == '0')
				n0++;
			if (s[i] == '1')
				n1++;
		}
		if (n0 > len-n0)
			puts ("00");
		if (n0 <= n1 && n0 >= len/2 || n0 > n1 && n1 >= (len+1)/2)
		{
			if (s[len-1] == '0')
				puts ("10");
			else if (s[len-1] == '1')
				puts ("01");
			else    //如果最后是?, 根据条件构造
			{
				if (n0-1 <= n1 && (n0-1) >= len/2 ||
					n0-1 > n1 && n1 >= (len+1)/2)
					puts ("01");
				if (n0 <= n1-1 && n0 >= len/2 ||
					n0 > n1-1 && (n1-1) >= (len+1)/2)
					puts ("10");
			}
		}
		if (n1 - (len-n1) > 1)
			puts ("11");
	}
    return 0;
}