SDOI2017 硬币游戏（数论）

题目链接

题目大意

给定一些长度相同的01串，接下来从第一位开始等概率随机01，直到出现某个给定的串为止。问每个串出现的概率是多少。

题解

神仙题orz……
直接判定两两之间的关系比较困难，考虑加入辅助变量 $x$表示当前没有任何串出现。
比如A=001,B=100
那么 $x$+011=A赢+(B赢+1)+(B赢+01)
即 $x/8=p_A+p_B/2+p_B/4$
于是对于每个变量都可以列出一个方程，现在有 $n$个方程。
可是有 $n+1$个变量，注意到 $\sum p_i=1$，因此直接高斯消元即可。

#include <bits/stdc++.h>
using namespace std;
const int MAXR = 10000000;
char _READ_[MAXR], _PRINT_[MAXR];
int _READ_POS_, _PRINT_POS_, _READ_LEN_;
inline char readc() {
#ifndef ONLINE_JUDGE
	return getchar();
#endif
	if (!_READ_POS_) _READ_LEN_ = fread(_READ_, 1, MAXR, stdin);
	char c = _READ_[_READ_POS_++];
	if (_READ_POS_ == MAXR) _READ_POS_ = 0;
	if (_READ_POS_ > _READ_LEN_) return 0;
	return c;
}
template<typename T> inline void read(T &x) {
	x = 0; register int flag = 1, c;
	while (((c = readc()) < '0' || c > '9') && c != '-');
	if (c == '-') flag = -1; else x = c - '0';
	while ((c = readc()) >= '0' && c <= '9') x = x * 10 + c - '0';
	x *= flag;
}
template<typename T1, typename ...T2> inline void read(T1 &a, T2&... x) {
	read(a), read(x...);
}
inline int reads(char *s) {
	register int len = 0, c;
	while (isspace(c = readc()) || !c);
	s[len++] = c;
	while (!isspace(c = readc()) && c) s[len++] = c;
	s[len] = 0;
	return len;
}
inline void ioflush() { fwrite(_PRINT_, 1, _PRINT_POS_, stdout), _PRINT_POS_ = 0; fflush(stdout); }
inline void printc(char c) {
	_PRINT_[_PRINT_POS_++] = c;
	if (_PRINT_POS_ == MAXR) ioflush();
}
inline void prints(char *s) {
	for (int i = 0; s[i]; i++) printc(s[i]);
}
template<typename T> inline void print(T x, char c = '\n') {
	if (x < 0) printc('-'), x = -x;
	if (x) {
		static char sta[20];
		register int tp = 0;
		for (; x; x /= 10) sta[tp++] = x % 10 + '0';
		while (tp > 0) printc(sta[--tp]);
	} else printc('0');
	printc(c);
}
template<typename T1, typename ...T2> inline void print(T1 x, T2... y) {
	print(x, ' '), print(y...);
}
typedef long long ll;

const int MAXN = 305;
const int EPS = 1E-9;
int nxt[MAXN * 2], n, m;
char str[MAXN][MAXN], temp[MAXN * 2];
double a[MAXN][MAXN], pw[MAXN];
void gauss(int n) {
	for (int i = 1; i <= n; i++) {
		int t = -1;
		for (int j = i; j <= n; j++)
			if (fabs(a[j][i]) > EPS) { t = j; break; }
		if (t != i) for (int j = i; j <= n; j++) swap(a[i][j], a[t][j]);
		for (int j = i + 1; j <= n; j++) {
			if (fabs(a[j][i]) < EPS) continue;
			double p = a[j][i] / a[i][i];
			for (int k = i; k <= n + 1; k++) a[j][k] -= p * a[i][k];
		}
	}
	for (int i = n; i > 0; i--) {
		for (int j = i + 1; j <= n; j++) {
			a[i][n + 1] -= a[i][j] * a[j][n + 1];
			a[i][j] = 0;
		}
		a[i][n + 1] /= a[i][i], a[i][i] = 1;
	}
}
int main() {
	read(n, m);
	for (int i = 1; i <= n; i++) reads(str[i]);
	pw[0] = 1;
	for (int i = 1; i <= m; i++) pw[i] = pw[i - 1] * 0.5;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			int cnt = 0;
			for (int k = 0; k < m; k++) temp[cnt++] = str[i][k];
			temp[cnt++] = '$';
			for (int k = 0; k < m; k++) temp[cnt++] = str[j][k];
			nxt[0] = -1;
			for (int k = 1, l = 0; k < cnt; k++) {
				while (l >= 0 && temp[l] != temp[k]) l = nxt[l];
				nxt[k + 1] = ++l;
			}
			double sum = 0;
			for (int k = nxt[cnt]; k; k = nxt[k]) sum += pw[m - k];
			a[i][j] = sum;
		}
		a[i][n + 1] = -pw[m];
		a[n + 1][i] = 1;
	}
	a[n + 1][n + 2] = 1;
	gauss(n + 1);
	for (int i = 1; i <= n; i++) printf("%.10lf\n", a[i][n + 2]);
	return 0;
}