topcoder srm 540 div1

problem1 link

设第一个数字为$x$，那么第２到第$n$个数字都可以表示成$a+bx$的形式，其中$b=1$或者$b=-1$．然后可以求出关于$x$的一些范围，求交集即可．

problem2 link

设$f[i][r][g][b]$表示前$i$个已经染色完毕，第$i$颜色为$(r,g,b)$的概率．假设由$(r,g,b)$可以转移到的状态个数为$T$,那么每个可以转移到的状态$(r^{'},g^{'},b^{'})$都可以得到$\frac{f[i][r][g][b]}{T}$的贡献．暴力转移的复杂度太高．如果没有$d_{1}$的限制，那么$r,g,b$可以转移到的状态，如果将其看作三维空间中的坐标，是以$(r,g,b)$为中心的边长为$2d_{2}+1$的立方体．这样的更新可以整块进行．有了$d_{1}$，可以将至少有一个大于等于$d_{1}$改为所有的情况减去都小于$d_{1}$的情况．这样的话，$r,g,b$可以转移到的状态为以$r,g,b$为中心的边长为$2d_{2}+1$的立方体去掉以$r,g,b$为中心的边长为$2d_{1}+1$的立方体．

problem3 link

首先，由于10=2*5．这样将问题分解为计算模２以及模５的种类数．然后乘起来即可．考虑计算这个子问题．令$p=2$或者$5$．

由于如果$output[i]=0$，那么对应的区间一定有至少一个０，反之，一定是一个０也不会有．那么可以将$N$个数字分成两部分，一定不会出现0的部分以及可能有０的部分．

（１）对于可能有０的部分的计算：设$f[i][j]$表示放置了前$i$个数字，最后一个放置０的位置在$j$的方案数．

（２）对于一定没有０的部分的计算：首先，如果两个不同的区间$[L_{1},R_{1}],[L_{2},R_{2}]$的关系是$L_{1}=L_{2}$或者$R_{1}=R_{2}$，那么可以将其中一个较大的区间强行分成两段．这样最后所有的区间的两两之间的关系一定以下三种：不相交，一个是另一个的真子集，相交但是不是真子集．如下图所示，有绿，红，蓝，黄，黑５个区间．对于每个区间来说，如果区间内前面的元素确定了，那么区间的最后一个元素就是唯一的（因为$p$是素数）．这样的话，只有问号的位置是可以随意放置的．

code for problem1

#include <vector>
#include <string>
#include <limits>
using namespace std;class ImportantSequence {public:int getCount(vector<int> B, string operators) {int64_t left = 1;int64_t right = std::numeric_limits<int64_t>::max();int64_t a = 0;int b = 1;for (size_t i = 0; i < B.size(); ++i) {int64_t val = static_cast<int64_t>(B[i]);if (operators[i] == '+') {a = val -a;b = -b;} else {a = a - val;}if (b == 1) {left = std::max(left, 1 - a);} else {right = std::min(right, a - 1);}if (left > right) {return 0;}}if (right == std::numeric_limits<int64_t>::max()) {return -1;}return static_cast<int>(right - left + 1);}
};

code for problem2

#include <algorithm>
#include <cstring>
#include <iostream>const int MAXN = 40;
const int MAX = 50;
double f[MAXN][MAX + 1][MAX + 1][MAX + 1];class RandomColoring {public:double getProbability(int N, int maxR, int maxG, int maxB, int startR,int startG, int startB, int d1, int d2) {this->max_r = maxR;this->max_g = maxG;this->max_b = maxB;f[0][startR][startG][startB] = 1;for (int i = 1; i < N; ++i) {for (int r = 0; r < maxR; ++r) {const std::pair<int, int> rd2 = ComputeRage(r, d2, maxR);const std::pair<int, int> rd1 = ComputeRage(r, d1 - 1, maxR);for (int g = 0; g < maxG; ++g) {const std::pair<int, int> gd2 = ComputeRage(g, d2, maxG);const std::pair<int, int> gd1 = ComputeRage(g, d1 - 1, maxG);for (int b = 0; b < maxB; ++b) {const std::pair<int, int> bd2 = ComputeRage(b, d2, maxB);const std::pair<int, int> bd1 = ComputeRage(b, d1 - 1, maxB);const int s1 = Size(rd2) * Size(gd2) * Size(bd2);const int s0 = d1 > 0 ? Size(rd1) * Size(gd1) * Size(bd1) : 0;if (s1 == s0) {continue;}const double e = f[i - 1][r][g][b] / (s1 - s0);Add(i, rd2.first, gd2.first, bd2.first, rd2.second, gd2.second,bd2.second, e);if (d1 > 0) {Add(i, rd1.first, gd1.first, bd1.first, rd1.second, gd1.second,bd1.second, -e);}}}}for (int r = 0; r < maxR; ++r) {for (int g = 0; g < maxG; ++g) {for (int b = 0; b < maxB; ++b) {f[i][r][g][b] += Get(i, r - 1, g, b) + Get(i, r, g - 1, b) +Get(i, r, g, b - 1) - Get(i, r - 1, g - 1, b) -Get(i, r - 1, g, b - 1) - Get(i, r, g - 1, b - 1) +Get(i, r - 1, g - 1, b - 1);}}}}double result = 0;for (int x = 0; x < maxR; ++x) {for (int y = 0; y < maxG; ++y) {for (int z = 0; z < maxB; ++z) {if (!(abs(x - startR) <= d2 && abs(y - startG) <= d2 &&abs(z - startB) <= d2 &&(abs(x - startR) >= d1 || abs(y - startG) >= d1 ||abs(z - startB) >= d1))) {result += f[N - 1][x][y][z];}}}}return result;}double Get(int t, int x, int y, int z) {if (x >= 0 && y >= 0 && z >= 0) {return f[t][x][y][z];}return 0.0;}void Add(int t, int x, int y, int z, int x1, int y1, int z1, double val) {Add(t, x, y, z, val);Add(t, x, y, z1 + 1, -val);Add(t, x, y1 + 1, z, -val);Add(t, x1 + 1, y, z, -val);Add(t, x1 + 1, y1 + 1, z, val);Add(t, x1 + 1, y, z1 + 1, val);Add(t, x, y1 + 1, z1 + 1, val);Add(t, x1 + 1, y1 + 1, z1 + 1, -val);}void Add(int t, int x, int y, int z, double val) {if (x < max_r && y < max_g && z < max_b) {f[t][x][y][z] += val;}}std::pair<int, int> ComputeRage(int x, int d, int max_x) {return {std::max(0, x - d), std::min(x + d, max_x - 1)};}int Size(const std::pair<int, int> &range) {return range.second - range.first + 1;}private:int max_r, max_g, max_b;
};

code for problem3

#include <string>
#include <vector>
#include <limits>
#include <iostream>
using namespace std;class ProductQuery {static const int64_t mod = 1000000007;public:int theInput(int N, vector<int> Qfrom, vector<int> Qto, vector<int> output) {auto ModFunc = [output](const int p) -> std::vector<int> {std::vector<int> r;for (auto x : output) {r.push_back(x % p);}return r;};return static_cast<int>(Compute(N, Qfrom, Qto, ModFunc(2), 2) *Compute(N, Qfrom, Qto, ModFunc(5), 5) %mod);}private:int64_t Compute(int N, const std::vector<int> &qfrom,const std::vector<int> &qto, const std::vector<int> &output,const int p) {std::vector<int> not_zero(N, 0);for (size_t i = 0; i < output.size(); ++i) {if (0 != output[i]) {for (int j = qfrom[i]; j <= qto[i]; ++j) {not_zero[j] = 1;}}}int zero_index = 0;int not_zero_index = 0;std::vector<int> zero_map(N, -1);std::vector<int> not_zero_map(N, -1);for (int i = 0; i < N; ++i) {if (not_zero[i] == 0) {zero_map[i] = zero_index++;} else {not_zero_map[i] = not_zero_index++;}}std::vector<int> zero_qfrom;std::vector<int> zero_qto;std::vector<int> not_zero_qfrom;std::vector<int> not_zero_qto;std::vector<int> not_zero_output;auto FindValidRange = [](const std::vector<int> &elements, int left,int right) -> std::pair<int, int> {int left_index = -1;int right_index = -1;for (int j = left; j <= right; ++j) {if (elements[j] != -1) {if (left_index == -1) {left_index = elements[j];}right_index = elements[j];}}return {left_index, right_index};};for (size_t i = 0; i < output.size(); ++i) {if (output[i] == 0) {auto r = FindValidRange(zero_map, qfrom[i], qto[i]);if (r.first == -1) {return 0;}zero_qfrom.push_back(r.first);zero_qto.push_back(r.second);} else {auto r = FindValidRange(not_zero_map, qfrom[i], qto[i]);if (r.first == -1) {return 0;}not_zero_qfrom.push_back(r.first);not_zero_qto.push_back(r.second);not_zero_output.push_back(output[i]);}}return Place0(zero_index, zero_qfrom, zero_qto, p) *NotPlace0(not_zero_index, not_zero_qfrom, not_zero_qto,not_zero_output, p) %mod;}int64_t Place0(int N, const std::vector<int> &qfrom,const std::vector<int> &qto, const int p) {if (N == 0) {return 1;}auto CheckForceZero = [qfrom, qto](int pre_zero_index,int cur_index) -> bool {for (size_t i = 0; i < qfrom.size(); ++i) {if (qfrom[i] > pre_zero_index && qto[i] == cur_index) {return true;}}return false;};auto EncodeIndex = [](int t) { return t + 1; };std::vector<std::vector<int64_t>> f(N, std::vector<int64_t>(N + 1, 0));f[0][EncodeIndex(0)] = 1;if (!CheckForceZero(-1, 0)) {f[0][EncodeIndex(-1)] = p - 1;}for (int i = 1; i < N; ++i) {for (int j = -1; j < i; ++j) {const int64_t val = f[i - 1][EncodeIndex(j)];Add(f[i][EncodeIndex(i)], val);if (!CheckForceZero(j, i)) {Add(f[i][EncodeIndex(j)], val * (p - 1) % mod);}}}int64_t result = 0;for (int i = -1; i < N; ++i) {Add(result, f[N - 1][EncodeIndex(i)]);}return result;}int64_t NotPlace0(int N, const std::vector<int> &qfrom,const std::vector<int> &qto, const std::vector<int> &output,const int p) {if (N == 0) {return 1;}std::vector<int64_t> rev_table(p);for (int i = 1; i < p; ++i) {for (int j = 1; j < p; ++j) {if (i * j % p == 1) {rev_table[i] = j;break;}}}std::vector<bool> delete_items(output.size(), false);std::vector<int> last_qfrom = qfrom;std::vector<int> last_qto = qto;std::vector<int> last_output = output;bool updated = true;int total_range = static_cast<int>(output.size());while (updated) {updated = false;for (size_t i = 0; i < output.size(); ++i) {if (!delete_items[i]) {for (int j = i + 1; j < output.size(); ++j) {if (!delete_items[j]) {if (last_qfrom[i] == last_qfrom[j] && last_qto[i] == last_qto[j]) {if (last_output[i] != last_output[j]) {return 0;}--total_range;delete_items[j] = true;updated = true;} else if (last_qfrom[i] == last_qfrom[j]) {if (last_qto[i] < last_qto[j]) {last_qfrom[j] = last_qto[i] + 1;last_output[j] =last_output[j] * rev_table[last_output[i]] % p;} else {last_qfrom[i] = last_qto[j] + 1;last_output[i] =last_output[i] * rev_table[last_output[j]] % p;}updated = true;} else if (last_qto[i] == last_qto[j]) {if (last_qfrom[i] < last_qfrom[j]) {last_qto[i] = last_qfrom[j] - 1;last_output[i] =last_output[i] * rev_table[last_output[j]] % p;} else {last_qto[j] = last_qfrom[i] - 1;last_output[j] =last_output[j] * rev_table[last_output[i]] % p;}updated = true;}}}}}}int64_t result = 1;for (int i = 1; i <= N - total_range; ++i) {result = result * (p - 1) % mod;}return result;}void Add(int64_t &x, int64_t y) {x += y;if (x >= mod) {x -= mod;}}
};

转载于:https://www.cnblogs.com/jianglangcaijin/p/8644956.html

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