topcoder srm 663 div1
problem1 link
每次枚举$S$的两种变化，并判断新的串是否是$T$的子串。不是的话停止搜索。
problem2 link
首先考慮增加１个面值为１的硬币后，$ways$数组有什么变化。设原来的方案为$w_{0}$，现在的为$w_{1}$。那么有$w_{1}[i]=w_{0}[i]+w_{0}[i-1]$。
$w_{0}: 1,5,6,7,0$
$w_{1}: 1,6,11,13,7$
如果将上面的例子中的$w_{0},w_{1}$看作多项式的话，就有$w_{0}=1+5x+6x^{2}+7x^{3}$,$w_{1}=1+6x+11x^{2}+13x^{3}+7x_{4}=(1+5x+6x^{2}+7x^{3})(1+x)=(1+x)w_{0}$
同理，面值为１的硬币增加$k$的话，就有$w_{1}[i]=\sum_{t=0}^{k}w_{0}[i-t]$，即多项式的话就是$w_{1}=(1+x)^{k}w_{0}$。如果是面值$v$的增加$k$的话，就是$w_{1}=(1+x^{v})^{k}w_{0}$。所以这里跟面值是１的是类似的。只分析面试为１的即可。
题目中是已知$w_{1}$来计算$w_{0}$，那么有$w_{0}=\frac{w_{1}}{(1+x)^{k}}$。
因为$ \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}...\rightarrow \frac{1}{(1+x)^{k}}=(1-x+x^{2}-x^{3}+x^{4}...)^{k}=\sum_{n\geq 0}(-1)^{n}C_{k+n-1}^{n}x^{n}$
所以如果$w_{1}=\sum_{i=0}^{D}a_{i}x^{i},w_{0}=\sum_{i=0}^{D}b_{i}x^{i}$
那么$b_{D}=\sum_{i\geq 0}(-1)^{i}C_{k+i-1}^{i}a_{D-i}$
如果是面值为$v$的增加$k$个的话就有$b_{D}=\sum_{i\geq 0}(-1)^{i}C_{k+i-1}^{i}a_{D-iv}$.
这个代码是$O(qD)$的复杂度。不过一直超时，不知道怎么回事。
problem3 link
将卡片以及操作看作是图的顶点和边。每个节点有一个颜色$A$或者$B$。首先，节点０的颜色永远不会变。最后所有节点的颜色都会变成节点０的颜色。假设$[1,n-1]$中与节点０相同颜色的节点有$x$个。将这些节点称为$good$
一个结论是在任意次操作后（直到结束前），所有$x$个$good$节点的分布情况是等概率的。
所以可以计算一开始有$0,1,2,...,n-1$个$good$节点的期望。设$i$个点的时候的期望为$f(i)$。那么经过一次操作后，$i$个$good$点有可能变为$i-1,i,i+1$个$good$点，设概率为$p_{1},p_{2},p_{3}$
其中$p_{1}=\frac{C_{i+1}^{2}+C_{n-(i+1)}^{2}}{C_{n}^{2}}$,$p_{2}=\frac{i(n-(i+1))}{2C_{n}^{2}},p_{3}=1-p_{1}-p_{2}$
转移方程为$f(i)=p_{1}f(i)+p_{2}f(i-1)+p_{3}f(i+1)+t$,
$t=\left\{\begin{matrix} 0 & i \neq s\\ \frac{1}{C_{n-1}^{s}} & i=s \end{matrix}\right.$
$s$为输入中$good$ 节点的个数。
求出$f$数组后，答案为$\frac{\sum_{i=0}^{n-1}f(i)C_{n-1}^{i}}{2^{n}}$
code for problem1
#include <algorithm>
#include <string>
#include <unordered_set>
#include <vector>
class ABBADiv1 {
public:
std::string canObtain(const std::string &a, const std::string &b) {
int n = static_cast<int>(b.size());
sub_sets.resize(n + 1);
for (int i = 0; i < n; ++i) {
for (int j = i; j < n; ++j) {
int len = j - i + 1;
std::string s = b.substr(i, len);
sub_sets[len].insert(s);
std::reverse(s.begin(), s.end());
sub_sets[len].insert(s);
}
}
if (Dfs(a, b)) {
return "Possible";
}
return "Impossible";
}
private:
std::vector<std::unordered_set<std::string>> sub_sets;
bool Exist(const std::string &s) { return sub_sets[s.size()].count(s) > 0; }
bool Dfs(const std::string &a, const std::string &target) {
if (a.size() == target.size()) {
return a == target;
}
if (Exist(a + 'A') && Dfs(a + 'A', target)) {
return true;
}
std::string new_a = a + 'B';
std::reverse(new_a.begin(), new_a.end());
return Exist(new_a) && Dfs(new_a, target);
}
};
code for problem2
#include <algorithm>
#include <vector>
constexpr int kMAXN = 1002000;
long long p[kMAXN];
long long q[kMAXN];
class ChangingChange {
public:
std::vector<int> countWays(std::vector<int> ways,
std::vector<int> valueRemoved,
std::vector<int> numRemoved) {
constexpr int kMod = 1000000007;
int n = ways.size() - 1;
int m = valueRemoved.size();
auto ModInv = [&](long long k) {
int t = kMod - 2;
long long r = 1;
while (t > 0) {
if (t % 2 == 1) {
r = r * k % kMod;
}
t /= 2;
k = k * k % kMod;
}
return r;
};
p[0] = 1;
for (int i = 1; i < kMAXN; ++i) {
p[i] = i * p[i - 1] % kMod;
}
q[kMAXN - 1] = ModInv(p[kMAXN - 1]);
for (int i = kMAXN - 2; i >= 0; --i) {
q[i] = q[i + 1] * (i + 1) % kMod;
}
std::vector<int> result(m);
for (int id = 0; id < m; ++id) {
int k = valueRemoved[id];
int num = numRemoved[id];
long long total = 0;
for (int i = 0, t = n / k; i <= t; ++i) {
long long a = p[num + i - 1] * q[i] % kMod * q[num - 1] % kMod *
ways[n - i * k] % kMod;
if ((i & 1) == 0) {
total += a;
} else {
total += kMod - a;
}
if (total >= kMod) {
total -= kMod;
}
}
result[id] = static_cast<int>(total);
}
return result;
}
};
code for problem3
#include <string>
constexpr int kMod = 1000000007;
constexpr int kMaxN = 1001;
int a[kMaxN][kMaxN];
int c[kMaxN][kMaxN];
int f[kMaxN];
class WarAndPeas {
public:
int expectedPeas(const std::string &state) {
int n = static_cast<int>(state.size());
c[0][0] = 1;
for (int i = 1; i <= n; ++i) {
c[i][0] = c[i][i] = 1;
for (int j = 1; j < i; ++j) {
c[i][j] = Add(c[i - 1][j - 1], c[i - 1][j]);
}
}
int s = 0;
for (int i = 1; i < n; ++i) {
if (state[i] == state[0]) {
++s;
}
}
for (int i = 0; i < n; ++i) {
int p1 = Mul(Add(c[i + 1][2], c[n - i - 1][2]), Inverse(c[n][2]));
int p2 = Mul(Mul(i, n - i - 1), Inverse(2 * c[n][2]));
int p3 = Add(1, kMod - Add(p1, p2));
if (i == n - 1) {
p1 = 0;
}
p1 = Add(p1, kMod - 1);
if (i == s) {
a[i][n] = Mul(kMod - 1, Inverse(c[n - 1][s]));
}
a[i][i] = p1;
if (i > 0) {
a[i][i - 1] = p2;
}
if (i < n - 1) {
a[i][i + 1] = p3;
}
}
Solve(n);
int result = 0;
for (int i = 0; i < n; ++i) {
result = Add(result, Mul(f[i], c[n - 1][i]));
}
result = Mul(result, Inverse(Pow(2, n)));
return result;
}
private:
void Solve(int n) {
for (int i = 0; i < n; ++i) {
for (int j = i; j < n; ++j) {
if (a[j][i] != 0) {
for (int k = 0; k <= n; ++k) {
std::swap(a[i][k], a[j][k]);
}
break;
}
}
for (int j = 0; j < n; ++j) {
if (j == i || a[j][i] == 0) {
continue;
}
int t = Mul(a[j][i], Inverse(a[i][i]));
for (int k = 0; k <= n; ++k) {
a[j][k] = Add(a[j][k], kMod - Mul(a[i][k], t));
}
}
}
for (int i = 0; i < n; ++i) {
f[i] = Mul(a[i][n], Inverse(a[i][i]));
}
}
int Pow(long long a, int b) {
long long r = 1;
while (b > 0) {
if (b % 2 == 1) {
r = r * a % kMod;
}
a = a * a % kMod;
b /= 2;
}
return static_cast<int>(r);
}
int Inverse(int x) { return Pow(x, kMod - 2); }
int Add(int x, int y) {
x += y;
if (x >= kMod) {
x -= kMod;
}
return x;
}
int Mul(long long x, long long y) { return static_cast<int>(x * y % kMod); }
};
