【spoj8757】Kmp 概率Dp

给出字符串S，S仅由a与b组成。有一个空串T，每次在末尾随机加上a或b。问期望几轮后S串为T串的后缀?

先用Kmp求出匹配失败去哪，然后可得状态转移。

f[i] = (sum[i-1] - sum[next[i] - 1] + f[i] + 1) * 0.5 + 0.5, f[0] = 2;

移项得 f[i] = sum[i-1] - sum[next[i] - 1] + 2。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <string>#define Rep(i, x, y) for (int i = x; i <= y; i ++)#define fi g[nx[i]][b[i+1]]using namespace std;typedef long long LL;const int N = 45;int T, n, nx[N], g[N][2];bool b[N];LL ans, f[N], sum[N];char s[N];void Next() {g[0][b[1]] = 1;Rep(i, 1, n) {g[i][b[i+1]] = i + 1;g[i][!b[i+1]] = g[nx[i]][!b[i+1]];nx[i+1] = fi;}}int main(){cin >> T;while (T --) {scanf ("%s", s+1);n = strlen(s+1);memset(nx, 0, sizeof(nx));memset(g, 0, sizeof(g));Rep(i, 1, n) b[i] = (s[i] == 'T');memset(f, 0, sizeof(f));Next();f[0] = sum[0] = 2;ans = 2;Rep(i, 1, n-1) {f[i] = sum[i-1] + 2;if (g[i][!b[i+1]]) f[i] -= sum[g[i][!b[i+1]]-1];sum[i] = sum[i-1] + f[i];ans += f[i];}cout << ans << endl;}        return 0;    }