UVaLive 3490 Generator (KMP + DP + Gauss)

题意：随机字母组成一个串，有一个目标串，当这个由随机字母组成的串出现目标串就停止，求这个随机字母组成串的期望长度。

析：由于只要包含目标串就可以停止，所以可以先把这个串进行处理，也就是KMP，然后dp[i] 表示从 i 结点到完全匹配期望长度，所以很容易得到状态转移方程 dp[i] = ∑dp[j] / n + 1，然后用高斯消元即可，要注意，要用全整数的高斯消元。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 150000 + 10;
const int maxm = 3e5 + 10;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[20];
int f[maxn];
LL A[20][20];

void getFail(int n){
  f[0] = f[1] = 0;
  for(int i = 1; i < n; ++i){
    int j = f[i];
    while(j && s[i] != s[j])  j = f[j];
    f[i+1] = s[i] == s[j] ? j+1 : 0;
  }
}

void Gauess(int n){
  for(int i = 0; i < n; ++i){
    int r = i;
    while(r < n && !A[r][i])  ++r;
    if(r != i)  for(int j = 0; j <= n; ++j)  swap(A[r][j], A[i][j]);

    for(int k = i+1; k < n; ++k) if(A[k][i]){
      LL f = A[k][i];
      for(int j = i; j <= n; ++j)  A[k][j] = A[k][j] * A[i][i] - f * A[i][j];
    }
  }
  for(int i = n-1; i >= 0; --i){
    for(int j = i+1; j < n; ++j)
      A[i][n] -= A[j][n] * A[i][j];
    A[i][n] /= A[i][i];
  }
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %s", &n, s);
    m = strlen(s);
    getFail(m);
    ms(A, 0);
    for(int i = 0; i < m; ++i){
      A[i][i] += n;
      A[i][m+1] += n;
      for(int k = 0; k < n; ++k){
        int j = i;
        while(j && s[j] != 'A' + k)  j = f[j];
        if(s[j] == 'A' + k)  ++j;
        --A[i][j];
      }
    }
    A[m][m] = 1;
    Gauess(m + 1);
    printf("Case %d:\n", kase);
    printf("%lld\n", A[0][m+1]);
    if(kase != T)  puts("");
  }
  return 0;
}