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;
}
}
}
}
}
}
