标签：get bsp highlight unity long lan link 不能 comm
problem1 link
如果decisions的大小为0，那么每一轮都是$N$个人。答案为0.
否则，如果答案不为0，那么概率最大的一定是一开始票数最多的人。因为这个人每一轮都在可以留下来的人群中。
假设第一轮之后剩下$r_{1}$个人，那么第二轮之后将剩下$r_{2}=N$%$r_{1}$个人。不妨设$r_{1}<N$。第一轮能够使得$r_{1}$个人的票数答案一样多且最多（设这个票数为$x$），那么这个x一定是大于等于单纯由decisions决定出的最大值。而现在$r_{1}<N$了，所以必然现在能投出的最大值还是大于等于第一轮的最大值。最后结果是：一些人的票数是$N/r_{1}$，另外一些是$N/r_{1}+1$，那么后一部分人将留下来到下一轮。
同理第$k$轮之后剩下$r_{k}=N$%$r_{k-1}$个人，直到剩下一个人。当出现$N$%$r_{i}=0$时，说明出现循环，答案为0。
有解的话，答案为$\frac{r_{2}}{r_{1}}*\frac{r_{3}}{r_{2}}*...*\frac{r_{k}}{r_{k-1}}=\frac{r_{k}}{r_{1}}=\frac{1}{r_{1}}$
problem2 link
首先，最后的图形的结果不会超出(-27,0)到(27,81)这个范围。
由于给出的范围都是整数，那么每次用给出的区间$R$与当前图形能扩展出的最大区间$R_{i}$比较，如果$R$能包含$R_{i}$，那么直接计算即可。否则可将当前线段继续增长，出现三个小区间，继续匹配即可。可以看出，最后最小的区间的大小为$1$x$1$。
对于一个线段$p_{1}p_{2}$，设还能增长$t$次，$L=|p_{1}p_{2}|$，那么最后的长度为$L(1+\frac{2}{3}t)$
problem3 link
$p_{5}$和$p_{7}$的个数就是5和7的个数。剩下的数字1，2，3，4，6，8，9出现的次数都不能确定。
可以枚举2,4,8出现的次数，这样根据$p_{2}$可以确定6出现的次数，再枚举9出现的次数，然后根据$p_{3}$以及6和9的个数可以确定3的个数，最后根据$S$可以确定1的个数。假设数字$i$出现的个数为$c_{i}$，令总位数$n=\sum_{i=1}^{9}c_{i}$。那么假设最后的某一位为数字$i$的方案数有$f(i)=\frac{(n-1)!}{c_{1}c_{2}...(c_{i}-1)...c_{8}c_{9}}$。然后数字$i$可以在第$0$位到第$n-1$位的任何一位，那么数字$i$对答案的贡献位$g(i)=f(i)*(1+10^{1}+...+10^{n-1})$
code for problem1
public class MafiaGame {
public double probabilityToLose(int N, int[] decisions) {
int[] c = new int[N];
for (int x : decisions) {
++ c[x];
}
int mx = 0;
for (int x : c) {
mx = Math.max(mx, x);
}
if (mx == 0) {
return 0;
}
int tot = N - decisions.length;
int r = 0;
for (int i = 0; i < N; ++ i) {
if (c[i] <= mx - 1) {
tot -= mx - 1 - c[i];
}
else {
r += 1;
}
}
if (tot > 0) {
r += tot;
}
double result = 1.0 / r;
while (r != 1) {
if (N % r == 0) {
return 0;
}
r = N % r;
}
return result;
}
}
code for problem2
public class FractalPicture {
static class Point {
int x, y;
Point() {}
Point(int x, int y) {
this.x = x;
this.y = y;
}
}
static class Rect {
int x1, x2, y1, y2;
Rect() {}
Rect(int x1, int y1, int x2, int y2) {
this.x1 = x1;
this.y1 = y1;
this.x2 = x2;
this.y2 = y2;
}
Rect intersect(Rect a) {
return new Rect(Math.max(x1, a.x1),
Math.max(y1, a.y1),
Math.min(x2, a.x2),
Math.min(y2, a.y2));
}
boolean contains(Rect r) {
return x1 <= r.x1 && r.x2 <= x2
&&y1 <= r.y1 && r.y2 <= y2;
}
}
static Rect extend(Point p1, Point p2) {
if (p1.x == p2.x) {
int y1 = Math.min(p1.y, p2.y);
int y2 = Math.max(p1.y, p2.y);
int detx = Math.max(1, (y2 - y1) / 3);
return new Rect(p1.x - detx, y1, p1.x + detx, y2);
}
else {
int x1 = Math.min(p1.x, p2.x);
int x2 = Math.max(p1.x, p2.x);
int dety = Math.max(1, (x2 - x1) / 3);
return new Rect(x1, p1.y - dety, x2, p1.y + dety);
}
}
static int length(Point p1, Point p2) {
if (p1.x == p2.x) {
return Math.abs(p1.y - p2.y);
}
return Math.abs(p1.x - p2.x);
}
public double getLength(int x1, int y1, int x2, int y2) {
Point p1 = new Point(0, 0);
Point p2 = new Point(0, 81);
Rect r = new Rect(x1, y1, x2, y2).intersect(extend(p1, p2));
return dfs(r, p1, p2, 499);
}
static int segIntersectSeg(int L1, int R1, int L2, int R2) {
return Math.max(0, Math.min(R1, R2) - Math.max(L1, L2));
}
static int rectIntersectSegment(Rect i, Point p1, Point p2) {
assert i.x1 <= i.x2 && i.y1 <= i.y2;
if (p1.x == p2.x) {
int x = p1.x;
int y1 = Math.min(p1.y, p2.y);
int y2 = Math.max(p1.y, p2.y);
if (i.x1 <= x && x <= i.x2) {
return segIntersectSeg(i.y1, i.y2, y1, y2);
}
else {
return 0;
}
}
else {
int y = p1.y;
int x1 = Math.min(p1.x, p2.x);
int x2 = Math.max(p1.x, p2.x);
if (i.y1 <= y && y <= i.y2) {
return segIntersectSeg(i.x1, i.x2, x1, x2);
}
else {
return 0;
}
}
}
static double calculate(Point p1, Point p2, int t) {
return length(p1, p2) * (1.0 + t * 2.0 / 3.0);
}
static double calculateUnit(Rect r, Point p1, Point p2, int t) {
double tot = calculate(p1, p2, t);
if (p1.x == p2.x) {
int x = p1.x;
int y1 = Math.min(p1.y, p2.y);
int y2 = Math.max(p1.y, p2.y);
if (r.x1 <= x && x <= r.x2 && r.y1 <= y1 && y2 <= r.y2) {
if (r.x1 == x || r.x2 == x) {
return (tot - 1) * 0.5 + 1;
}
return tot;
}
return 0;
}
else {
int y = p1.y;
int x1 = Math.min(p1.x, p2.x);
int x2 = Math.max(p1.x, p2.x);
if (r.y1 <= y && y <= r.y2 && r.x1 <= x1 && x2 <= r.x2) {
if (r.y1 == y || r.y2 == y) {
return (tot - 1) * 0.5 + 1;
}
return tot;
}
return 0;
}
}
static Point[] explode(Point p1, Point p2) {
if (p1.x == p2.x) {
int len = p2.y - p1.y;
Point pp1 = new Point(p1.x, p1.y + len / 3 * 2);
Point pp2 = new Point(p1.x - len / 3, pp1.y);
Point pp3 = new Point(p1.x + len / 3, pp1.y);
return new Point[]{pp1, pp2, pp3};
}
else {
int len = p2.x - p1.x;
Point pp1 = new Point(p1.x + len / 3 * 2, p1.y);
Point pp2 = new Point(pp1.x, p1.y - len / 3);
Point pp3 = new Point(pp1.x, p1.y + len / 3);
return new Point[]{pp1, pp2, pp3};
}
}
double dfs(Rect r, Point p1, Point p2, int t) {
if (r.x1 > r.x2 || r.y1 > r.y2) {
return 0;
}
if (t == 0) {
return rectIntersectSegment(r, p1, p2);
}
if (r.contains(extend(p1, p2))) {
return calculate(p1, p2, t);
}
if (length(p1, p2) == 1) {
return calculateUnit(r, p1, p2, t);
}
Point[] ps = explode(p1, p2);
double result = rectIntersectSegment(r, p1, ps[0]);
result += dfs(r, ps[0], ps[1], t - 1);
result += dfs(r, ps[0], ps[2], t - 1);
result += dfs(r, ps[0], p2, t - 1);
return result;
}
}
code for problem3
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class ProductAndSum {
final static int mod = 500500573;
public int getSum(int p2, int p3, int p5, int p7, int S) {
long[] p = new long[S + 1];
long[] q = new long[S + 1];
p[0] = q[0] = 1;
for (int i = 1; i <= S; ++ i) {
p[i] = p[i - 1] * i % mod;
q[i] = pow(p[i], mod - 2);
}
long[] f = new long[S + 1];
f[0] = 1;
for (int i = 1, b = 1; i <= S; ++ i) {
b = (int)(b * 10L % mod);
f[i] = (f[i - 1] + b) % mod;
}
int[] c = new int[10];
c[5] = p5;
c[7] = p7;
long result = 0;
for (c[2] = 0; c[2] <= p2; ++ c[2]) {
for (c[4] = 0; c[4] * 2 + c[2] <= p2; ++ c[4]) {
for (c[8] = 0; c[8] * 3 + c[4] * 2 + c[2] <= p2; ++ c[8]) {
c[6] = p2 - c[2] - c[4] * 2 - c[8] * 3;
for (c[9] = 0; c[9] * 2 +c[6] <= p3; ++ c[9]) {
c[3] = p3 - c[6] - c[9] * 2;
c[1] = S;
for (int i = 2; i <= 9; ++ i) {
c[1] -= c[i] * i;
}
if (c[1] < 0) {
continue;
}
int n = 0;
for (int i = 1; i <= 9; ++ i) {
n += c[i];
}
long k = p[n - 1];
for (int i = 1; i <= 9; ++ i) {
k = k * q[c[i]] % mod;
}
for (int i = 1; i <= 9; ++ i) {
if (c[i] != 0) {
result += k * p[c[i]] % mod * q[c[i] - 1] % mod * f[n - 1] % mod * i % mod;
result %= mod;
}
}
}
}
}
}
return (int)result;
}
static long pow(long a, long b) {
a %= mod;
long result = 1;
while (b > 0) {
if (b % 2 == 1) {
result = result * a % mod;
}
a = a * a % mod;
b >>= 1;
}
return result;
}
}
标签：get bsp highlight unity long lan link 不能 comm
原文地址：http://www.cnblogs.com/jianglangcaijin/p/7906349.html
