topcoder srm 475 div1

problem1 link

暴力枚举$r$只兔子的初始位置,然后模拟即可。

problem2 link

假设刚生下来的兔子是1岁,那么能够生小兔子的兔子的年龄是至少3岁。

那么所有的兔子按照年龄可以分为1岁,2岁,大于等于3岁三种情况。不妨令个数分别为$a_{1},a_{2},a_{3}$

在每年生完兔子后,假如是四月,这时候1岁的兔子和大于等于3岁的兔子的数量是一样的($a_{1}=a_{3}$)。那么在11月的时候,如果要去掉一半的兔子,那么一定是大于等于三岁的所有兔子和2岁兔子的一半。$a_{3}=0,a_{2}=frac{a_{2}}{2}$ or $a_{2}=frac{a_{2}+1}{2}$

现在的问题是要确定现在2岁的兔子的个数是奇数还是偶数。不妨令三种兔子的奇偶数分别是$x_{1},x_{2},x_{3}$

假设$x_{2}=t*2^{51}+r$,$0 leq r < 2^{51}$

那么只需要用$r$的奇偶性来代表$x_{2}$的奇偶性即可,即便有除以2,以及加1再除以2的操作。因为$t*2^{51}$除以若干次2后(小于等于50次)仍然是偶数。

problem3 link

假设有$n$只兔子

首先计算出每个兔子得分的最大值最小值。minPoints,maxPoints

设选出的selected只兔子的最后一只的编号为$x$, 假设它此时的得分 maxPoints[x]

另外假设其他选择的兔子都是它们的最高得分,而未选兔子都是它们的最低得分

令$f[i][j][k]$表示在前$i$只兔子中,有$j$只的排名高于$x$且在这$j$只中选出了$k$只的情况有多少种。那此次的答案为$Answer_{x}=sum_{i= selected}^{qualified}f[n][i][selected]$

这里需要考虑的是,会不会出现这样的情况导致少计算了方案数:由于$x$固定得了maxPoints[x]分,而如果它的得分少于maxPoints[x]时,可能会出现某个兔子$t$的排名在$x$之前。(比如$minPoints[x]<maxPoints[t]<maxPoints[x]$)

其实这种情况不会漏算,因为当枚举$t$作为最后选出兔子的最后一只时,会算上这种情况。

因此,只需要考虑选择的兔子都是它们的最高得分,而未选兔子都是它们的最低得分即可。

code for problem1

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class RabbitStepping {

	int result = 0;
	String field = null;
	int[] q = null;
	int[] pre = null;

	public double getExpected(String field, int r) {
		this.field = field;
		q = new int[field.length()];
		pre = new int[field.length()];
		dfs(0, 0, r, new int[field.length()]);
		long p = 1;
		for (int i = 1; i <= r; ++ i) {
			p = p * (field.length() + 1 - i) / i;
		}
		return 1.0 * result / p;
	}
	void dfs(int dep, int start, int tot, int[] p) {
		if (dep == tot) {
			cal(p);
			return;
		}
		if (start == p.length) {
			return;
		}
		if (tot - dep > p.length - start) {
			return;
		}
		for (int i = start; i < p.length; ++ i) {
			p[i] = 1;
			dfs(dep + 1, i + 1, tot, p);
			p[i] = 0;
		}
	}

	void move(int from, int to) {
		q[to] += 1;
		q[from] -= 1;
		pre[to] = from;
	}

	void cal(int[] p) {
		for (int i = 0; i < p.length; ++ i) {
			q[i] = p[i];
		}
		int size = q.length;
		Arrays.fill(pre, -1);
		int[] pre1 = new int[p.length];
		int[] q1 = new int[p.length];
		while (size > 2) {
			for (int i = 0; i < size; ++ i) {
				pre1[i] = pre[i];
				q1[i] = q[i];
			}
			for (int i = 0; i < size; ++ i) {
				if (q1[i] == 0) {
					continue;
				}
				if (i == 0) {
					move(0, 1);
				}
				else if (i == size - 1 || i == size - 2) {
					move(i, i - 1);
				}
				else if (field.charAt(i) == 'W'){
					move(i, i - 1);
				}
				else if (field.charAt(i) == 'B') {
					move(i, i + 1);
				}
				else {
					if (size == p.length) {
						move(i, i - 1);
					}
					else {
						move(i, pre1[i]);
					}
				}
			}
			for (int i = 0; i < size; ++ i) {
				if (q[i] >= 2) {
					q[i] = 0;
					pre[i] = 0;
				}
			}
			size -= 1;
		}
		for (int i = 0; i < size; ++ i) {
			result += q[i];
		}
	}
}

  


code for problem2

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class RabbitIncreasing {

	final static int mod = 1000000009;
	final static long MOD = 1l << 51;
	final static int rev2 = Pow(2, mod - 2, mod);

	public int getNumber(int[] leaving, int k) {
		int a3 = 0, a2 = 0, a1 = 1;
		long x3 = 0, x2 = 0, x1 = 1;
		if (leaving[0] == 1) {
		    return 0;
		}
		int t = 0;
		for (int i = 2; i <= k; ++ i) {
			a3 = (a3 + a2) % mod;
			a2 = a1;
			a1 = a3;
			x3 = (x3 + x2) % MOD;
			x2 = x1;
			x1 = x3;
			if (t < leaving.length && leaving[t] == i) {
				x3 = 0;
				a3 = 0;
				if ((x2 & 1) == 1) {
					x2 -= 1;
					a2 = (a2 + mod - 1) % mod;
				}
				x2 >>= 1;
				a2 = (int)((long)a2 * rev2 % mod);
				t += 1;
			}
		}
		return ((a3 + a2) % mod + a1) % mod;

	}

	static int Pow(int a, int b, int mod) {
		int result = 1;
		a %= mod;
		while (b != 0) {
			if ((b & 1) == 1) {
				result = (int)((long)result * a % mod);
			}
			a = (int)((long)a * a % mod);
			b >>= 1;
		}
		return result;
	}


}

  


code for problem3

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class RabbitProgramming {

	int[] minPoints = null;
	int[] maxPoints = null;
	long[][][] f = null;
	int n;
	int qualified;
	int selected;

	public long getTeams(int[] points, String[] standings, int qualified, int selected) {
		this.n = standings.length;
		this.qualified = qualified;
		this.selected = selected;
		minPoints = new int[n];
		maxPoints = new int[n];
		for (int i = 0; i < points.length; ++ i) {
			for (int j = 0; j < n; ++ j) {
				if (standings[j].charAt(i) == 'N') {
					continue;
				}
				if (points[i] > 0) {
					minPoints[j] += points[i];
					maxPoints[j] += points[i];
				}
				else {
					maxPoints[j] += -points[i];
				}
			}
		}
		f = new long[n][qualified + 1][selected + 1];
		long result = 0;
		for (int x = 0; x < n; ++ x) {
			cal(x);
			for (int i = selected; i <= qualified; ++ i) {
				result += f[n - 1][i][selected];
			}
		}
		return result;
	}
	void cal(int x) {
		for (int i = 0; i < n; ++ i) {
			for (int j = 0; j <= qualified; ++ j) {
				Arrays.fill(f[i][j], 0);
			}
		}
		if (x == 0) {
			f[0][1][1] = 1;
		}
		else {
			if (minPoints[0] < maxPoints[x]) {
				f[0][0][0] = 1;
			}
			else {
				f[0][1][0] = 1;
			}
			if (maxPoints[0] >= maxPoints[x]) {
				f[0][1][1] = 1;
			}
		}
		for (int i = 1; i < n; ++ i) {
			for (int j = 0; j <= qualified; ++ j) {
				for (int k = 0; k <= selected; ++ k) {
					final long val = f[i - 1][j][k];
					if(val == 0) {
						continue;
					}
					if (i == x) {
						if (j < qualified && k < selected) {
							f[i][j + 1][k + 1] += val;
						}
						continue;
					}
					if (i < x && minPoints[i] < maxPoints[x]
							|| i > x && minPoints[i] <= maxPoints[x]) {
						f[i][j][k] += val;
					}
					else if(j < qualified) {
						f[i][j + 1][k] += val;
					}
					if ((i < x && maxPoints[i] >= maxPoints[x]
							|| i > x && maxPoints[i] > maxPoints[x])
							 && j < qualified && k <selected) {
						f[i][j + 1][k + 1] += val;
					}
				}
			}
		}
	}
}

  

原文地址:https://www.cnblogs.com/jianglangcaijin/p/7821194.html